The origin of the name “Napkin” comes from the following quote of mine.
I’ll be eating a quick lunch with some friends of mine who are still in high school.They’ll ask me what I’ve been up to the last few weeks, and I’ll tell them thatI’ve been learning category theory. They’ll ask me what category theory is about.I tell them it’s about abstracting things by looking at just the structure-preservingmorphisms between them, rather than the objects themselves. I’ll try to give themthe standard exampleGrp, but then I’ll realize that they don’t know what ahomomorphism is. So then I’ll start trying to explain what a homomorphism is,but then I’ll remember that they haven’t learned what a group is. So then I’ll starttrying to explain what a group is, but by the time I finish writing the group axiomson my napkin, they’ve already forgotten why I was talking about groups in the firstplace. And then it’s 1PM, people need to go places, and I can’t help but think: “Man, if I had forty hours instead of forty minutes, I bet I could actually haveexplained this all”.
This book was initially my attempt at those forty hours, but has grown considerably since
then.
About this book
The Infinitely Large Napkin is a light but mostly self-contained introduction to a large amount
of higher math.
I should say at once that this book is not intended as a replacement for dedicated books or
courses; the amount of depth is not comparable. On the flip side, the benefit of this “light”
approach is that it becomes accessible to a larger audience, since the goal is merely to give the
reader a feeling for any particular topic rather than to emulate a full semester of
lectures.
I initially wrote this book with talented high-school students in mind, particularly those
with math-olympiad type backgrounds. Some remnants of that cultural bias can still be felt
throughout the book, particularly in assorted challenge problems which are taken from
mathematical competitions. However, in general I think this would be a good reference for
anyone with some amount of mathematical maturity and curiosity. Examples include but
certainly not limited to: math undergraduate majors, physics/CS majors, math PhD students
who want to hear a little bit about fields other than their own, advanced high schoolers
who like math but not math contests, and unusually intelligent kittens fluent in
English.
Source code
The project is hosted on GitHub at https://github.com/vEnhance/napkin. Pull
requests are quite welcome! I am also happy to receive suggestions and corrections by
email.
Philosophy behind the Napkin approach
As far as I can tell, higher math for high-school students comes in two flavors:
Someone tells you about the hairy ball theorem in the form “you can’t comb the
hair on a spherical cat” then doesn’t tell you anything about why it should be
true, what it means to actually “comb the hair”, or any of the underlying theory,
leaving you with just some vague notion in your head.
You take a class and prove every result in full detail, and at some point you stop
caring about what the professor is saying.
Presumably you already know how unsatisfying the first approach is. So the second approach
seems to be the default, but I really think there should be some sort of middle ground
here.
Unlike university, it is not the purpose of this book to train you to solve exercises or write
proofs1 , or
prepare you for research in the field. Instead I just want to show you some interesting
math. The things that are presented should be memorable and worth caring about.
For that reason, proofs that would be included for completeness in any ordinary
textbook are often omitted here, unless there is some idea in the proof which I think is
worth seeing. In particular, I place a strong emphasis over explaining why a theorem
should be true rather than writing down its proof. This is a recurrent theme of this
book:
Natural explanations supersede proofs.
My hope is that after reading any particular chapter in Napkin, one might get the following
out of it:
Knowing what the precise definitions are of the main characters,
Being acquainted with the few really major examples,
Knowing the precise statements of famous theorems, and having a sense of why
they should be true.
Understanding “why” something is true can have many forms. This is sometimes accomplished
with a complete rigorous proof; in other cases, it is given by the idea of the proof; in still other
cases, it is just a few key examples with extensive cheerleading.
Obviously this is nowhere near enough if you want to e.g. do research in a field; but if you
are just a curious outsider, I hope that it’s more satisfying than the elevator pitch or
Wikipedia articles. Even if you do want to learn a topic with serious depth, I hope
that it can be a good zoomed-out overview before you really dive in, because in
many senses the choice of material is “what I wish someone had told me before I
started”.
More pedagogical comments and references
The preface would become too long if I talked about some of my pedagogical decisions chapter
by chapter, so ?? contains those comments instead.
In particular, I often name specific references, and the end of that appendix has more
references. So this is a good place to look if you want further reading.
Historical and personal notes
I began writing this book in December of 2014, after having finished my first semester of
undergraduate at Harvard. It became my main focus for about 18 months after that, as I
became immersed in higher math. I essentially took only math classes, (gleefully ignoring all
my other graduation requirements) and merged as much of it as I could (as well as lots of
other math I learned on my own time) into the Napkin.
Towards August of 2016, though, I finally lost steam. The first public drafts
went online then, and I decided to step back. Having burnt out slightly, I then
took a break from higher math, and spent the remaining two undergraduate
years2
working extensively as a coach for the American math olympiad team, and trying to spend as
much time with my friends as I could before they graduated and went their own
ways.
During those two years, readers sent me many kind words of gratitude, many reports of
errors, and many suggestions for additions. So in November of 2018, some weeks into my first
semester as a math PhD student, I decided I should finish what I had started. Some months
later, here is what I have.
Acknowledgements
I am indebted to countless people for this work. Here is a partial (surely incomplete)
list.
Thanks to all my teachers and professors for teaching me much of the material
covered in these notes, as well as the authors of all the references I have cited here.
A special call-out to [?], [?], [?], [?], [?], [?], [?], [?], [?], [?], which were especially
influential.
Thanks also to dozens of friends and strangers who read through preview copies
of my draft, and pointed out errors and gave other suggestions. Special mention
to Andrej Vuković and Alexander Chua for together catching over a thousand
errors. Thanks also to Brian Gu and Tom Tseng for many corrections. (If you find
mistakes or have suggestions yourself, I would love to hear them!)
I’d also like to express my gratitude for many, many kind words I received during
the development of this project. These generous comments led me to keep working
on this, and were largely responsible for my decision in November 2018 to begin
updating the Napkin again.
Finally, a huge thanks to the math olympiad community, from which the Napkin (and me)
has its roots. All the enthusiasm, encouragement, and thank-you notes I have received over the
years led me to begin writing this in the first place. I otherwise would never have the
arrogance to dream a project like this was at all possible. And of course I would be nowhere
near where I am today were it not for the life-changing journey I took in chasing my dreams to
the IMO. Forever TWN2!
Advice for the reader
1Prerequisites
As explained in the preface, the main prerequisite is some amount of mathematical maturity.
This means I expect the reader to know how to read and write a proof, follow logical
arguments, and so on.
I also assume the reader is familiar with basic terminology about sets and functions
(e.g. “what is a bijection?”). If not, one should consult ?? .
2Deciding what to read
There is no need to read this book in linear order: it covers all sorts of areas in mathematics,
and there are many paths you can take. In ?? , I give a short overview for each part
explaining what you might expect to see in that part.
For now, here is a brief chart showing how the chapters depend on each other; again see ??
for details. Dependencies are indicated by arrows; dotted lines are optional dependencies. Isuggest that you simply pick a chapter you find interesting, and then find theshortest path. With that in mind, I hope the length of the entire PDF is not
intimidating.
3Questions, exercises, and problems
In this book, there are three hierarchies:
An inline question is intended to be offensively easy, mostly a chance to help you
internalize definitions. If you find yourself unable to answer one or two of them, it
probably means I explained it badly and you should complain to me. But if you
can’t answer many, you likely missed something important: read back.
An inline exercise is more meaty than a question, but shouldn’t have any “tricky”
steps. Often I leave proofs of theorems and propositions as exercises if they are
instructive and at least somewhat interesting.
Each chapter features several trickier problems at the end. Some are
reasonable, but others are legitimately difficult olympiad-style problems.
Harder problems are marked with up to three chili peppers ( ),
like this paragraph.
In addition to difficulty annotations, the problems are also marked by how important
they are to the big picture.
Normal problems, which are hopefully fun but non-central.
Daggered problems, which are (usually interesting) results that one should
know, but won’t be used directly later.
Starred problems, which are results which will be used later on in the
book.1
Several hints and solutions can be found in ?? .
Image from [?]
4Paper
At the risk of being blunt,
Read this book with pencil and paper.
Here’s why:
Image from [?]
You are not God. You cannot keep everything in your
head.2
If you’ve printed out a hard copy, then write in the margins. If you’re trying to save paper,
grab a notebook or something along with the ride. Somehow, some way, make sure you can
write. Thanks.
5On the importance of examples
I am pathologically obsessed with examples. In this book, I place all examples in large boxes
to draw emphasis to them, which leads to some pages of the book simply consisting of
sequences of boxes one after another. I hope the reader doesn’t mind.
I also often highlight a “prototypical example” for some sections, and reserve the color red
for such a note. The philosophy is that any time the reader sees a definition or a theorem
about such an object, they should test it against the prototypical example. If the
example is a good prototype, it should be immediately clear why this definition is
intuitive, or why the theorem should be true, or why the theorem is interesting, et
cetera.
Let me tell you a secret. Whenever I wrote a definition or a theorem in this book, I would
have to recall the exact statement from my (quite poor) memory. So instead I often consider
the prototypical example, and then only after that do I remember what the definition or the
theorem is. Incidentally, this is also how I learned all the definitions in the first place. I hope
you’ll find it useful as well.
6Conventions and notations
This part describes some of the less familiar notations and definitions and settles for once and
for all some annoying issues (“is zero a natural number?”). Most of these are “remarks for
experts”: if something doesn’t make sense, you probably don’t have to worry about it for
now.
A full glossary of notation used can be found in the appendix.
6.iNatural numbers are positive
The set ℕ is the set of positive integers, not including 0. In the set theory chapters, we use
ω = {0,1,…} instead, for consistency with the rest of the book.
6.iiSets and equivalence relations
This is brief, intended as a reminder for experts. Consult ?? for full details.
An equivalence relation on a set X is a relation ∼ which is symmetric, reflexive, and
transitive. An equivalence relation partitions X into several equivalence classes. We will
denote this by X∕∼. An element of such an equivalence class is a representative of that
equivalence class.
I always use for an “isomorphism”-style relation (formally: a relation which is an
isomorphism in a reasonable category). The only time ≃ is used in the Napkin is for
homotopic paths.
I generally use ⊆ and ⊊ since these are non-ambiguous, unlike ⊂. I only use ⊂ on rare
occasions in which equality obviously does not hold yet pointing it out would be distracting.
For example, I write ℚ ⊂ ℝ since “ℚ ⊊ ℝ” is distracting.
I prefer S ∖ T to S − T.
The power set of S (i.e., the set of subsets of S), is denoted either by 2S or 𝒫(S).
6.iiiFunctions
This is brief, intended as a reminder for experts. Consult ?? for full details.
Let XY be a function:
By fpre(T) I mean the pre-image
This is in contrast to the f−1(T) used in the rest of the world; I only use f−1 for
an inverse function.
By abuse of notation, we may abbreviate fpre({y}) to fpre(y). We call fpre(y) a
fiber.
By fimg(S) I mean the image
Almost everyone else in the world uses f(S) (though f[S] sees some use, and f“(S)
is often used in logic) but this is abuse of notation, and I prefer fimg(S) for
emphasis. This image notation is not standard.
If S ⊆ X, then the restriction of f to S is denoted f↾S, i.e. it is the function
f↾S : S → Y .
Sometimes functions f : X → Y are injective or surjective; I may emphasize this
sometimes by writing f : XY or f : X ↠ Y , respectively.
6.ivCycle notation for permutations
Additionally, a permutation on a finite set may be denoted in cycle notation, as
described in say https://en.wikipedia.org/wiki/Permutation#Cycle_notation. For
example the notation (1 2 3 4)(5 6 7) refers to the permutation with 12, 23,
34, 41, 56, 67, 75. Usage of this notation will usually be obvious from
context.
6.vRings
All rings have a multiplicative identity 1 unless otherwise specified. We allow 0 = 1 in general
rings but not in integral domains.
All rings are commutative unless otherwise specified. There is an elaborate scheme
for naming rings which are not commutative, used only in the chapter on cohomology
rings:
Graded
Not Graded
1 not required
graded pseudo-ring
pseudo-ring
Anticommutative, 1 not required
anticommutative pseudo-ring
N/A
Has 1
graded ring
N/A
Anticommutative with 1
anticommutative ring
N/A
Commutative with 1
commutative graded ring
ring
On the other hand, an algebra always has 1, but it need not be commutative.
6.viChoice
We accept the Axiom of Choice, and use it freely.
7Further reading
The appendix ?? contains a list of resources I like, and explanations of pedagogical choices
that I made for each chapter. I encourage you to check it out.
In particular, this is where you should go for further reading! There are some topics that
should be covered in the Napkin, but are not, due to my own ignorance or laziness. The
references provided in this appendix should hopefully help partially atone for my
omissions.
This chapter contains a pitch for each part, to help you decide what you want to read and to
elaborate more on how they are interconnected.
For convenience, here is again the dependency plot that appeared in the frontmatter.
0.1The basics
Starting Out. I made a design decision that the first part should have a little bit both of algebra
and topology: so this first chapter begins by defining a group, while the second
chapter begins by defining a metric space. The intention is so that newcomers
get to see two different examples of “sets with additional structure” in somewhat
different contexts, and to have a minimal amount of literacy as these sorts of
definitions appear over and over.1
Basic Abstract Algebra. The algebraically inclined can then delve into further types of algebraic structures:
some more details of groups, and then rings and fields — which will let you
generalize ℤ, ℚ, ℝ, ℂ. So you’ll learn to become familiar with all sorts of other
nouns that appear in algebra, unlocking a whole host of objects that one couldn’t
talk about before.
We’ll also come into ideals, which generalize the GCD in ℤ that you might know
of. For example, you know in ℤ that any integer can be written in the form 3a+5b
for a,b ∈ ℤ, since gcd(3,5) = 1. We’ll see that this statement is really a statement
of ideals: “(3,5) = 1 in ℤ”, and thus we’ll understand in what situations it can be
generalized, e.g. to polynomials.
Basic Topology. The more analytically inclined can instead move into topology, learning more
about spaces. We’ll find out that “metric spaces” are actually too specific, and
that it’s better to work with topological spaces, which are based on the so-called
open sets. You’ll then get to see the buddings of some geometrical ideals, ending
with the really great notion of compactness, a powerful notion that makes real
analysis tick.
One example of an application of compactness to tempt you now: a continuous
function f : [0,1] → ℝ always achieves a maximum value. (In contrast, f : (0,1) →ℝ by x1∕x does not.) We’ll see the reason is that [0,1] is compact.
0.2Abstract algebra
Linear Algebra. In high school, linear algebra is often really unsatisfying. You are given these
arrays of numbers, and they’re manipulated in some ways that don’t really make
sense. For example, the determinant is defined as this funny-looking sum with a
bunch of products that seems to come out of thin air. Where does it come from?
Why does det(AB) = detAdetB with such a bizarre formula?
Well, it turns out that you can explain all of these things! The trick is to not think
of linear algebra as the study of matrices, but instead as the study of linear maps.
In earlier chapters we saw that we got great generalizations by speaking of “sets
with enriched structure” and “maps between them”. This time, our sets are vectorspaces and our maps are linear maps. We’ll find out that a matrix is actually
just a way of writing down a linear map as an array of numbers, but using the
“intrinsic” definitions we’ll de-mystify all the strange formulas from high school
and show you where they all come from.
In particular, we’ll see easy proofs that column rank equals row rank, determinant
is multiplicative, trace is the sum of the diagonal entries. We’ll see how the dot
product works, and learn all the words starting with “eigen-”. We’ll even have a
bonus chapter for Fourier analysis showing that you can also explain all the big
buzz-words by just being comfortable with vector spaces.
More on Groups. Some of you might be interested in more about groups, and this chapter will give
you a way to play further. It starts with an exploration of group actions, then
goes into a bit on Sylow theorems, which are the tools that let us try to classifyall groups.
Representation Theory. If G is a group, we can try to understand it by implementing it as a matrix,
i.e. considering embeddings GGLn(ℂ). These are called representations of
G; it turns out that they can be decomposed into irreducible ones. Astonishingly
we will find that we can basically characterize all of them: the results turn out to
be short and completely unexpected.
For example, we will find out that there are finitely many irreducible
representations of a given finite group G; if we label them V1, V2, …, Vr, then we
will find that r is the number of conjugacy classes of G, and moreover that
which comes out of nowhere!
The last chapter of this part will show you some unexpected corollaries. Here is
one of them: let G be a finite group and create variables xg for each g ∈ G. A
|G|×|G| matrix M is defined by setting the (g,h)th entry to be the variable xg⋅h.
Then this determinant will turn out to factor, and the factors will correspond to
the Vi we described above: there will be an irreducible factor of degree dimVi
appearing dimVi times. This result, called the Frobenius determinant, is said
to have given birth to representation theory.
Quantum Algorithms. If you ever wondered what Shor’s algorithm is, this chapter will use the built-up
linear algebra to tell you!
0.3Real and complex analysis
Calculus 101. In this part, we’ll use our built-up knowledge of metric and topological spaces to
give short, rigorous definitions and theorems typical of high school calculus. That
is, we’ll really define and prove most everything you’ve seen about limits, series,
derivatives, and integrals.
Although this might seem intimidating, it turns out that actually, by the time
we start this chapter, the hard work has already been done: the notion of limits,
open sets, and compactness will make short work of what was swept under the rug
in AP calculus. Most of the proofs will thus actually be quite short. We sit back
and watch all the pieces slowly come together as a reward for our careful study of
topology beforehand.
That said, if you are willing to suspend belief, you can actually read most of the
other parts without knowing the exact details of all the calculus here, so in some
sense this part is “optional”.
Complex Analysis. It turns out that holomorphic functions (complex-differentiable functions) are close
to the nicest things ever: they turn out to be given by a Taylor series (i.e. are basically
polynomials). This means we’ll be able to prove unreasonably nice results about
holomorphic functions ℂ → ℂ, like
they are determined by just a few inputs,
their contour integrals are all zero,
they can’t be bounded unless they are constant,
….
We then introduce meromorphic functions, which are like quotients of holomorphic
functions, and find that we can detect their zeros by simply drawing loops in the plane
and integrating over them: the famous residue theorem appears. (In the practice
problems, you will see this even gives us a way to evaluate real integrals that can’t be
evaluated otherwise.)
Measure Theory. Measure theory is the upgraded version of integration. The Riemann integration is
for a lot of purposes not really sufficient; for example, if f is the function
equals 1 at rational numbers but 0 at irrational numbers, we would hope that
∫01f(x) dx = 0, but the Riemann integral is not capable of handling this function
f.
The Lebesgue integral will handle these mistakes by assigning a measure to a generic
space Ω, making it into a measure space. This will let us develop a richer theory of
integration where the above integral does work out to zero because the “rational
numbers have measure zero”. Even the development of the measure will be an
achievement, because it means we’ve developed a rigorous, complete way of talking
about what notions like area and volume mean — on any space, not just ℝn! So for
example the Lebesgue integral will let us integrate functions over any measurespace.
Probability (TO DO). Using the tools of measure theory, we’ll be able to start giving rigorous definitions of
probability, too. We’ll see that a random variable is actually a function from a
measure space of worlds to ℝ, giving us a rigorous way to talk about its probabilities. We
can then start actually stating results like the law of large numbers and centrallimit theorem in ways that make them both easy to state and straightforward to
prove.
Differential Geometry. Multivariable calculus is often confusing because of all the partial derivatives.
But we’ll find out that, armed with our good understanding of linear algebra,
that we’re really looking at a total derivative: at every point of a function
f : ℝn→ ℝ we can associate a linear map Df which captures in one object
the notion of partial derivatives. Set up this way, we’ll get to see versions of
differential forms and Stokes’ theorem, and we finally will know what the
notation dx really means. In the end, we’ll say a little bit about manifolds in
general.
0.4Algebraic number theory
Algebraic NT I: Rings of Integers. Why is 3+ the conjugate of 3−? How come the norm = a2−5b2
used in Pell equations just happens to be multiplicative? Why is it we can do
factoring into primes in ℤ[i] but not in ℤ[]? All these questions and more will
be answered in this part, when we learn about number fields, a generalization
of ℚ and ℤ to things like ℚ() and ℤ[]. We’ll find out that we have unique
factorization into prime ideals, that there is a real multiplicative norm in play here,
and so on. We’ll also see that Pell’s equation falls out of this theory.
Algebraic NT II: Galois and Ramification Theory. All the big buzz-words come out now: Galois groups, the Frobenius, and friends.
We’ll see quadratic reciprocity is just a shadow of the behavior of the Frobenius
element, and meet the Chebotarev density theorem, which generalizes greatly
the Dirichlet theorem on the infinitude of primes which are a (mod n). Towards
the end, we’ll also state Artin reciprocity, one of the great results of class fieldtheory, and how it generalizes quadratic reciprocity and cubic reciprocity.
0.5Algebraic topology
Algebraic Topology I: Homotopy. What’s the difference between an annulus and disk? Well, one of them has a
“hole” in it, but if we are just given intrinsic topological spaces it’s hard to make
this notion precise. The fundamental groupπ1(X) and more general homotopygroup will make this precise — we’ll find a way to define an abelian group π1(X)
for every topological space X which captures the idea there is a hole in the space,
by throwing lassos into the space and seeing if we can reel them in.
Amazingly, the fundamental group π1(X) will, under mild conditions, tell you
about ways to cover X with a so-called covering projection. One picture is that
one can wrap a real line ℝ into a helix shape and then project it down into the
circle S1. This will turn out to correspond to the fact that π1(S1) = ℤ which has
only one subgroup. More generally the subgroups of π1(X) will be in bijection with
ways to cover the space X!
Category Theory. What do fields, groups, manifolds, metric spaces, measure spaces, modules,
representations, rings, topological spaces, vector spaces, all have in common?
Answer: they are all “objects with additional structure”, with maps between them.
The notion of category will appropriately generalize all of them. We’ll see that all
sorts of constructions and ideas can be abstracted into the framework of a category,
in which we only think about objects and arrows between them, without probing
too hard into the details of what those objects are. This results in drawing many
commutative diagrams.
For example, any way of taking an object in one category and getting another one
(for example π1 as above, from the category of spaces into the category of groups)
will probably be a functor. We’ll unify G×H, X ×Y , R ×S, and anything with
the × symbol into the notion of a product, and then even more generally into a
limit. Towards the end, we talk about abelian categories and talk about the
famous snake lemma, five lemma, and so on.
Algebraic Topology II: Homology. Using the language of category theory, we then resume our adventures in algebraic
topology, in which we define the homology groups which give a different way of
noticing holes in a space, in a way that is longer to define but easier to compute in
practice. We’ll then reverse the construction to get so-called cohomology rings
instead, which give us an even finer invariant for telling spaces apart.
0.6Algebraic geometry
Algebraic Geometry I: Classical Varieties. We begin with a classical study of classical complex varieties: the study of
intersections of polynomial equations over ℂ. This will naturally lead us into the
geometry of rings, giving ways to draw pictures of ideals, and motivating Hilbert’snullstellensatz. The Zariski topology will show its face, and then we’ll play
with projective varities and quasi-projective varieties, with a bonus detour
into Bezout’s theorem. All this prepares us for our journey into schemes.
Algebraic Geometry II: Affine Schemes. We now get serious and delve into Grothendiek’s definition of an affine scheme:
a generalization of our classical varieties that lets us start with any ring A and
construct a space SpecA on it. We’ll equip it with its own Zariski topology and
then a sheaf of functions on it, making it into a locally ringed space; we will find
that the sheaf can be understood effectively in terms of localization on it. We’ll
find that the language of commutative algebra provides elegant generalizations
of what’s going on geometrically: prime ideals correspond to irreducible closed
subsets, radical ideals correspond to closed subsets, maximal ideals correspond to
closed points, and so on. We’ll draw lots of pictures of spaces and examples to
accompany this.
Set Theory I: ZFC, Ordinals, and Cardinals. Why is Russell’s paradox such a big deal and how is it resolved? What is this
Zorn’s lemma that everyone keeps talking about? In this part we’ll learn the
answers to these questions by giving a real description of the Zermelo-Frankel
axioms, and the axiom of choice, delving into the details of how math is built
axiomatically at the very bottom foundations. We’ll meet the ordinal numbers
and cardinal numbers and learn how to do transfinite induction with them.
Set Theory II: Model Theory and Forcing. The continuum hypothesis states that there are no cardinalities between the
size of the natural numbers and the size of the real numbers. It was shown to be
independent of the axioms — one cannot prove or disprove it. How could a result
like that possibly be proved? Using our understanding of the ZF axioms, we’ll
develop a bit of model theory and then use forcing in order to show how to
construct entire models of the universe in which the continuum hypothesis is true
or false.
1Groups
A group is one of the most basic structures in higher mathematics. In this chapter I will
tell you only the bare minimum: what a group is, and when two groups are the
same.
1.1Definition and examples of groups
Prototypical example for this section:The additive group of integers (ℤ,+) and the cyclicgroup ℤ∕mℤ. Just don’t let yourself forget that most groups are non-commutative.
A group consists of two pieces of data: a set G, and an associative binary operation ⋆
with some properties. Before I write down the definition of a group, let me give two
examples.
Example 1.1.1 (Additive integers) The pair (ℤ,+) is a group: ℤ = is
the set and the associative operation is addition. Note that
The element 0 ∈ ℤ is an identity: a + 0 = 0 + a = a for any a.
Every element a ∈ ℤ has an additive inverse: a + (−a) = (−a) + a = 0.
We call this group ℤ.
Example 1.1.2 (Nonzero rationals) Let ℚ× be the set of nonzero rational numbers. The
pair (ℚ×,⋅) is a group: the set is ℚ× and the associative operation is multiplication.
Again we see the same two nice properties.
The element 1 ∈ ℚ× is an identity: for any rational number, a ⋅ 1 = 1 ⋅ a = a.
For any rational number x ∈ ℚ×, we have an inverse x−1, such that
From this you might already have a guess what the definition of a group is.
Definition 1.1.3.A group is a pair G = (G,⋆) consisting of a set of elements G, and a
binary operation ⋆ on G, such that:
G has an identity element, usually denoted 1G or just 1, with the property that
The operation is associative, meaning (a ⋆ b) ⋆ c = a ⋆ (b ⋆ c) for any a,b,c ∈ G.
Consequently we generally don’t write the parentheses.
Each element g ∈ G has an inverse, that is, an element h ∈ G such that
Remark 1.1.4 (Unimportant pedantic point) — Some authors like to add a “closure”
axiom, i.e. to say explicitly that g ⋆ h ∈ G. This is implied already by the fact that ⋆ is
a binary operation on G, but is worth keeping in mind for the examples below.
Remark 1.1.5 — It is not required that ⋆ is commutative (a⋆b = b⋆a). So we say that
a group is abelian if the operation is commutative and non-abelian otherwise.
Example 1.1.6 (Non-Examples of groups)
The pair (ℚ,⋅) is NOT a group. (Here ℚ is rational numbers.) While there is an
identity element, the element 0 ∈ ℚ does not have an inverse.
The pair (ℤ,⋅) is also NOT a group. (Why?)
Let Mat2×2(ℝ) be the set of 2 × 2 real matrices. Then (Mat2×2(ℝ),⋅) (where ⋅ is
matrix multiplication) is NOT a group. Indeed, even though we have an identity
matrix
we still run into the same issue as before: the zero matrix does not have a
multiplicative inverse.
(Even if we delete the zero matrix from the set, the resulting structure is still not
a group: those of you that know some linear algebra might recall that any matrix
with determinant zero cannot have an inverse.)
Let’s resume writing down examples. Here are some more abelian examples of
groups:
Example 1.1.7 (Complex unit circle) Let S1 denote the set of complex numbers z with
absolute value one; that is
Then (S1,×) is a group because
The complex number 1 ∈ S1 serves as the identity, and
Each complex number z ∈ S1 has an inverse which is also in S1, since
= −1 = 1.
There is one thing I ought to also check: that z1× z2 is actually still in S1. But this follows
from the fact that = = 1.
Example 1.1.8 (Addition mod n) Here is an example from number theory: Let n > 1
be an integer, and consider the residues (remainders) modulo n. These form a group
under addition. We call this the cyclic group of order n, and denote it as ℤ∕nℤ, with
elements 0,1,…. The identity is 0.
Example 1.1.9 (Multiplication mod p) Let p be a prime. Consider the nonzero residuesmodulo p, which we denote by (ℤ∕pℤ)×. Then is a group.
Question 1.1.10.Why do we need the fact that p is prime?
(Digression: the notation ℤ∕nℤ and (ℤ∕pℤ)× may seem strange but will make sense when
we talk about rings and ideals. Set aside your worry for now.)
Here are some non-abelian examples:
Example 1.1.11 (General linear group) Let n be a positive integer. Then GLn(ℝ) is
defined as the set of n × n real matrices which have nonzero determinant. It turns out
that with this condition, every matrix does indeed have an inverse, so (GLn(ℝ),×) is a
group, called the general linear group.
(The fact that GLn(ℝ) is closed under × follows from the linear algebra fact that
det(AB) = detAdetB, proved in later chapters.)
Example 1.1.12 (Special linear group) Following the example above, let SLn(ℝ) denote
the set of n × n matrices whose determinant is actually 1. Again, for linear algebra
reasons it turns out that (SLn(ℝ),×) is also a group, called the special linear group.
Example 1.1.13 (Symmetric groups) Let Sn be the set of permutations of . By
viewing these permutations as functions from to itself, we can consider
compositions of permutations. Then the pair (Sn,∘) (here ∘ is function composition) is also a
group, because
There is an identity permutation, and
Each permutation has an inverse.
The group Sn is called the symmetric group on n elements.
Example 1.1.14 (Dihedral group) The dihedral group of order 2n, denoted D2n,
is the group of symmetries of a regular n-gon A1A2…An, which includes rotations and
reflections. It consists of the 2n elements
The element r corresponds to rotating the n-gon by , while s corresponds to reflecting
it across the line OA1 (here O is the center of the polygon). So rs mean “reflect then
rotate” (like with function composition, we read from right to left).
In particular, rn = s2 = 1. You can also see that rks = sr−k.
Here is a picture of some elements of D10.
Trivia: the dihedral group D12 is my favorite example of a non-abelian group, and is the first
group I try for any exam question of the form “find an example…”.
More examples:
Example 1.1.15 (Products of groups) Let (G,⋆) and (H,∗) be groups. We can define a
product group (G × H,⋅), as follows. The elements of the group will be ordered pairs
(g,h) ∈ G × H. Then
is the group operation.
Question 1.1.16.What are the identity and inverses of the product group?
Example 1.1.17 (Trivial group) The trivial group, often denoted 0 or 1, is the group
with only an identity element. I will use the notation {1}.
Exercise 1.1.18.Which of these are groups?
(a)
Rational numbers with odd denominators (in simplest form), where the operation is addition.(This includes integers, written as n∕1, and 0 = 0∕1).
(b)
The set of rational numbers with denominator at most 2, where the operation is addition.
(c)
The set of rational numbers with denominator at most 2, where the operation ismultiplication.
(d)
The set of nonnegative integers, where the operation is addition.
1.2Properties of groups
Prototypical example for this section: (ℤ∕pℤ)×is possibly best.
Abuse of Notation 1.2.1.From now on, we’ll often refer to a group (G,⋆) by just G.
Moreover, we’ll abbreviate a ⋆ b to just ab. Also, because the operation ⋆ is associative,
we will omit unnecessary parentheses: (ab)c = a(bc) = abc.
Abuse of Notation 1.2.2.From now on, for any g ∈ G and n ∈ ℕ we abbreviate
Moreover, we let g−1 denote the inverse of g, and g−n = (g−1)n.
In mathematics, a common theme is to require that objects satisfy certain minimalistic
properties, with certain examples in mind, but then ignore the examples on paper and try to
deduce as much as you can just from the properties alone. (Math olympiad veterans are likely
familiar with “functional equations” in which knowing a single property about a function is
enough to determine the entire function.) Let’s try to do this here, and see what we can
conclude just from knowing ?? .
It is a law in Guam and 37 other states that I now state the following proposition.
Fact 1.2.3.Let G be a group.
(a)
The identity of a group is unique.
(b)
The inverse of any element is unique.
(c)
For any g ∈ G, (g−1)−1 = g.
Proof.This is mostly just some formal manipulations, and you needn’t feel bad skipping it on
a first read.
(a)
If 1 and 1′ are identities, then 1 = 1 ⋆ 1′ = 1′.
(b)
If h and h′ are inverses to g, then 1G = g ⋆ hh′ = (h′ ⋆ g) ⋆ h = 1G⋆ h = h.
(c)
Trivial; omitted. □
Now we state a slightly more useful proposition.
Proposition 1.2.4 (Inverse of products) Let G be a group, and a,b ∈ G. Then (ab)−1 =
b−1a−1.
Finally, we state a very important lemma about groups, which highlights why having an
inverse is so valuable.
Lemma 1.2.5 (Left multiplication is a bijection) Let G be a group, and pick a g ∈ G. Then
the map G → G given by xgx is a bijection.
Exercise 1.2.6.Check this by showing injectivity and surjectivity directly. (If you don’t knowwhat these words mean, consult?? .)
Example 1.2.7 Let G = (ℤ∕7ℤ)× (as in ?? ) and pick g = 3. The above lemma states that
the map x3 ⋅ x is a bijection, and we can see this explicitly:
1
3 (mod 7)
2
6 (mod 7)
3
2 (mod 7)
4
5 (mod 7)
5
1 (mod 7)
6
4 (mod 7).
The fact that the map is injective is often called the cancellation law. (Why do you think
so?)
Abuse of Notation 1.2.8 (Later on, sometimes the identity is denoted 0 instead of 1).You don’t need to worry about this for a few chapters, but I’ll bring it up now anyways.
In most of our examples up until now the operation ⋆ was thought of like multiplication
of some sort, which is why 1 = 1G was a natural notation for the identity element.
But there are groups like ℤ = (ℤ,+) where the operation ⋆ is thought of as addition,
in which case the notation 0 = 0G might make more sense instead. (In general, whenever
an operation is denoted +, the operation is almost certainly commutative.) We will
eventually start doing so too when we discuss rings and linear algebra.
1.3Isomorphisms
Prototypical example for this section:ℤ10ℤ.
First, let me talk about what it means for groups to be isomorphic. Consider the two groups
ℤ = (,+).
10ℤ = (,+).
These groups are “different”, but only superficially so – you might even say they only differ
in the names of the elements. Think about what this might mean formally for a
moment.
Specifically the map
is a bijection of the underlying sets which respects the group action. In symbols,
In other words, ϕ is a way of re-assigning names of the elements without changing the
structure of the group. That’s all just formalism for capturing the obvious fact that (ℤ,+) and
(10ℤ,+) are the same thing.
Now, let’s do the general definition.
Definition 1.3.1.Let G = (G,⋆) and H = (H,∗) be groups. A bijection ϕ : G → H is
called an isomorphism if
If there exists an isomorphism from G to H, then we say G and H are isomorphic
and write GH.
Note that in this definition, the left-hand side ϕ(g1⋆ g2) uses the operation of G while the
right-hand side ϕ(g1) ∗ ϕ(g2) uses the operation of H.
Example 1.3.2 (Examples of isomorphisms) Let G and H be groups. We have the following
isomorphisms.
(a)
ℤ10ℤ, as above.
(b)
There is an isomorphism
by the map (g,h)(h,g).
(c)
The identity map id : G → G is an isomorphism, hence GG.
(d)
There is another isomorphism of ℤ to itself: send every x to −x.
Example 1.3.3 (Primitive roots modulo 7) As a nontrivial example, we claim that
ℤ∕6ℤ(ℤ∕7ℤ)×. The bijection is
To check that this is an isomorphism, we need to verify several things.
First, we need to check this map actually makes sense: why is it the case that if
a ≡ b (mod 6), then 3a≡ 3b (mod 7)? The reason is that Fermat’s little theorem
guarantees that 36≡ 1 (mod 7).
Next, we need to check that this map is a bijection. You can do this explicitly:
Finally, we need to verify that this map respects the group action. In other words,
we want to see that ϕ(a+b) = ϕ(a)ϕ(b) since the operation of ℤ∕6ℤ is addition while
the operation of (ℤ∕7ℤ)× is multiplication. That’s just saying that 3a+b≡ 3a3b
(mod 7), which is true.
Example 1.3.4 (Primitive roots) More generally, for any prime p, there exists an
element g ∈ (ℤ∕pℤ)× called a primitive root modulo p such that 1,g,g2,…,gp−2 are
all different modulo p. One can show by copying the above proof that
The example above was the special case p = 7 and g = 3.
Exercise 1.3.5.Assuming the existence of primitive roots, establish the isomorphism ℤ∕(p −1)ℤ(ℤ∕pℤ)×as above.
It’s not hard to see that is an equivalence relation (why?). Moreover, because we really
only care about the structure of groups, we’ll usually consider two groups to be the same when
they are isomorphic. So phrases such as “find all groups” really mean “find all groups up to
isomorphism”.
1.4Orders of groups, and Lagrange’s theorem
Prototypical example for this section: (ℤ∕pℤ)×.
As is typical in math, we use the word “order” for way too many things. In groups, there are
two notions of order.
Definition 1.4.1.The order of a groupG is the number of elements of G. We denote
this by . Note that the order may not be finite, as in ℤ. We say G is a finite group
just to mean that is finite.
Example 1.4.2 (Orders of groups) For a prime p, = p − 1. In other words,
the order of (ℤ∕pℤ)× is p−1. As another example, the order of the symmetric group Sn
is = n! and the order of the dihedral group D2n is 2n.
Definition 1.4.3.The order of an elementg ∈ G is the smallest positive integer n
such that gn = 1G, or ∞ if no such n exists. We denote this by ordg.
Example 1.4.4 (Examples of orders) The order of −1 in ℚ× is 2, while the order of 1
in ℤ is infinite.
Question 1.4.5.Find the order of each of the six elements of ℤ∕6ℤ, the cyclic group on sixelements. (See?? if you’ve forgotten what ℤ∕6ℤ means.)
Example 1.4.6 (Primitive roots) If you know olympiad number theory, this coincides
with the definition of an order of a residue mod p. That’s why we use the term “order”
there as well. In particular, a primitive root is precisely an element g ∈ (ℤ∕pℤ)× such
that ordg = p − 1.
You might also know that if xn≡ 1 (mod p), then the order of x (mod p) must divide n.
The same is true in a general group for exactly the same reason.
Fact 1.4.7.If gn = 1G then ordg divides n.
Also, you can show that any element of a finite group has a finite order. The proof is just an
olympiad-style pigeonhole argument. Consider the infinite sequence 1G,g,g2,…, and find two
elements that are the same.
Fact 1.4.8.Let G be a finite group. For any g ∈ G, ordg is finite.
What’s the last property of (ℤ∕pℤ)× that you know from olympiad math? We have Fermat’s
little theorem: for any a ∈ (ℤ∕pℤ)×, we have ap−1≡ 1 (mod p). This is no coincidence: exactly
the same thing is true in a more general setting.
Theorem 1.4.9 (Lagrange’s theorem for orders) Let G be any finite group. Then x = 1G
for any x ∈ G.
Keep this result in mind! We’ll prove it later in the generality of ?? .
1.5Subgroups
Prototypical example for this section: SLn(ℝ) is a subgroup of GLn(ℝ).
Earlier we saw that GLn(ℝ), the n×n matrices with nonzero determinant, formed a group
under matrix multiplication. But we also saw that a subset of GLn(ℝ), namely SLn(ℝ),
also formed a group with the same operation. For that reason we say that SLn(ℝ)
is a subgroup of GLn(ℝ). And this definition generalizes in exactly the way you
expect.
Definition 1.5.1.Let G = (G,⋆) be a group. A subgroup of G is exactly what you
would expect it to be: a group H = (H,⋆) where H is a subset of G. It’s a propersubgroup if H≠G.
Remark 1.5.2 — To specify a group G, I needed to tell you both what the set G was
and the operation ⋆ was. But to specify a subgroup H of a given group G, I only need
to tell you who its elements are: the operation of H is just inherited from the operation
of G.
Example 1.5.3 (Examples of subgroups)
(a)
2ℤ is a subgroup of ℤ, which is isomorphic to ℤ itself!
(b)
Consider again Sn, the symmetric group on n elements. Let T be the set of
permutations τ : {1,…,n}→{1,…,n} for which τ(n) = n. Then T is a subgroup of
Sn; in fact, it is isomorphic to Sn−1.
(c)
Consider the group G × H (?? ) and the elements . This is a
subgroup of G × H (why?). In fact, it is isomorphic to G by the isomorphism
(g,1H)g.
Example 1.5.4 (Stupid examples of subgroups) For any group G, the trivial group {1G}
and the entire group G are subgroups of G.
Next is an especially important example that we’ll talk about more in later
chapters.
Example 1.5.5 (Subgroup generated by an element) Let x be an element of a group G.
Consider the set
This is also a subgroup of G, called the subgroup generated by x.
Exercise 1.5.6.Ifordx = 2015, what is the above subgroup equal to? What ifordx = ∞?
Finally, we present some non-examples of subgroups.
Example 1.5.7 (Non-examples of subgroups) Consider the group ℤ = (ℤ,+).
(a)
The set is not a subgroup of ℤ because it does not contain inverses.
(b)
The set {n3∣n ∈ ℤ} = {…,−8,−1,0,1,8,…} is not a subgroup because it is not closed
under addition; the sum of two cubes is not in general a cube.
(c)
The empty set ∅ is not a subgroup of ℤ because it lacks an identity element.
1.6Groups of small orders
Just for fun, here is a list of all groups of order less than or equal to ten (up to isomorphism,
of course).
1.
The only group of order 1 is the trivial group.
2.
The only group of order 2 is ℤ∕2ℤ.
3.
The only group of order 3 is ℤ∕3ℤ.
4.
The only groups of order 4 are
ℤ∕4ℤ, the cyclic group on four elements,
ℤ∕2ℤ × ℤ∕2ℤ, called the Klein Four Group.
5.
The only group of order 5 is ℤ∕5ℤ.
6.
The groups of order six are
ℤ∕6ℤ, the cyclic group on six elements.
S3, the permutation group of three elements. This is the first non-abelian
group.
Some of you might wonder where ℤ∕2ℤ × ℤ∕3ℤ is. All I have to say is: Chinese
remainder theorem!
You might wonder where D6 is in this list. It’s actually isomorphic to S3.
7.
The only group of order 7 is ℤ∕7ℤ.
8.
The groups of order eight are more numerous.
ℤ∕8ℤ, the cyclic group on eight elements.
ℤ∕4ℤ × ℤ∕2ℤ.
ℤ∕2ℤ × ℤ∕2ℤ × ℤ∕2ℤ.
D8, the dihedral group with eight elements, which is not abelian.
A non-abelian group Q8, called the quaternion group. It consists of eight
elements ±1, ±i, ±j, ±k with i2 = j2 = k2 = ijk = −1.
9.
The groups of order nine are
ℤ∕9ℤ, the cyclic group on nine elements.
ℤ∕3ℤ × ℤ∕3ℤ.
10.
The groups of order 10 are
ℤ∕10ℤℤ∕5ℤ × ℤ∕2ℤ (again Chinese remainder theorem).
D10, the dihedral group with 10 elements. This group is non-abelian.
1.7Unimportant long digression
A common question is: why these axioms? For example, why associative but not commutative?
This answer will likely not make sense until later, but here are some comments that may
help.
One general heuristic is: Whenever you define a new type of general object, there’s always a
balancing act going on. On the one hand, you want to include enough constraints that your
objects are “nice”. On the other hand, if you include too many constraints, then your
definition applies to too few objects.
So, for example, we include “associative” because that makes our lives easier and most
operations we run into are associative. In particular, associativity is required for the inverse of
an element to necessarily be unique. However we don’t include “commutative”,
because examples below show that there are lots of non-abelian groups we care about.
(But we introduce another name “abelian” because we still want to keep track of
it.)
Another comment: a good motivation for the inverse axioms is that you get a large amount
of symmetry. The set of positive integers with addition is not a group, for example, because
you can’t subtract 6 from 3: some elements are “larger” than others. By requiring an inverse
element to exist, you get rid of this issue. (You also need identity for this; it’s hard to define
inverses without it.)
Even more abstruse comment: ?? shows that groups are actually shadows of the so-called
symmetric groups (defined later, also called permutation groups). This makes rigorous the
notion that “groups are very symmetric”.
1.8A few harder problems to think about
Problem 1A.What is the joke in the following figure? (Source: [?].)
Problem 1B.Prove Lagrange’s theorem for orders in the special case that G is a finite
abelian group.
Problem 1C.Show that D6S3 but D24≇S4.
Problem 1D⋆.Let p be a prime. Show that the only group of order p is ℤ∕pℤ.
Problem 1E (A hint for Cayley’s theorem).Find a subgroup H of S8 which is
isomorphic to D8, and write the isomorphism explicitly.
Problem 1F†.Let G be a finite
group.1
Show that there exists a positive integer n such that
(a)
(Cayley’s theorem) G is isomorphic to some subgroup of the symmetric group Sn.
(b)
(Representation Theory) G is isomorphic to some subgroup of the general linear
group GLn(ℝ). (This is the group of invertible n × n matrices.)
Problem 1G (IMO SL 2005 C5).There are n markers, each with
one side white and the other side black. In the beginning, these n markers are aligned in
a row so that their white sides are all up. In each step, if possible, we choose a marker
whose white side is up (but not one of the outermost markers), remove it, and reverse
the closest marker to the left of it and also reverse the closest marker to the right of it.
Prove that if n ≡ 1 (mod 3) it’s impossible to reach a state with only two markers
remaining. (In fact the converse is true as well.)
Problem 1H.Let p be a prime and F1 = F2 = 1, Fn+2 = Fn+1 + Fn be
the Fibonacci sequence. Show that F2p(p2−1) is divisible by p.
2Metric spaces
At the time of writing, I’m convinced that metric topology is the morally
correct way to motivate point-set topology as well as to generalize normal
calculus.1
So here is my best attempt.
The concept of a metric space is very “concrete”, and lends itself easily to visualization.
Hence throughout this chapter you should draw lots of pictures as you learn about new
objects, like convergent sequences, open sets, closed sets, and so on.
2.1Definition and examples of metric spaces
Prototypical example for this section:ℝ2, with the Euclidean metric.
Definition 2.1.1.A metric space is a pair (M,d) consisting of a set of points M and a
metricd : M × M → ℝ≥0. The distance function must obey:
For any x,y ∈ M, we have d(x,y) = d(y,x); i.e. d is symmetric.
The function d must be positive definite which means that d(x,y) ≥ 0 with
equality if and only if x = y.
The function d should satisfy the triangle inequality: for all x,y,z ∈ M,
Abuse of Notation 2.1.2.Just like with groups, we will abbreviate (M,d) as just M.
Example 2.1.3 (Metric spaces of ℝ)
(a)
The real line ℝ is a metric space under the metric d(x,y) = .
(b)
The interval [0,1] is also a metric space with the same distance function.
(c)
In fact, any subset S of ℝ can be made into a metric space in this way.
Example 2.1.4 (Metric spaces of ℝ2)
(a)
We can make ℝ2 into a metric space by imposing the Euclidean distance function
(b)
Just like with the first example, any subset of ℝ2 also becomes a metric space after
we inherit it. The unit disk, unit circle, and the unit square [0,1]2 are special cases.
Example 2.1.5 (Taxicab on ℝ2) It is also possible to place the taxicab distance on
ℝ2:
For now, we will use the more natural Euclidean metric.
Example 2.1.6 (Metric spaces of ℝn) We can generalize the above examples easily. Let n be a
positive integer.
(a)
We let ℝn be the metric space whose points are points in n-dimensional Euclidean
space, and whose metric is the Euclidean metric
This is the n-dimensional Euclidean space.
(b)
The open unit ballBn is the subset of ℝn consisting of those points
such that x12 + + xn2< 1.
(c)
The unit sphereSn−1 is the subset of ℝn consisting of those points
such that x12 + + xn2 = 1, with the inherited metric. (The superscript
n − 1 indicates that Sn−1 is an n − 1 dimensional space, even though it lives in
n-dimensional space.) For example, S1⊆ ℝ2 is the unit circle, whose distance
between two points is the length of the chord joining them. You can also think of it
as the “boundary” of the unit ball Bn.
Example 2.1.7 (Function space) We can let M be the space of continuous functions
f : [0,1] → ℝ and define the metric by d(f,g) = ∫01dx. (It admittedly takes
some work to check d(f,g) = 0 implies f = g, but we won’t worry about that yet.)
Here is a slightly more pathological example.
Example 2.1.8 (Discrete space) Let S be any set of points (either finite or infinite). We
can make S into a discrete space by declaring
If = 4 you might think of this space as the vertices of a regular tetrahedron, living
in ℝ3. But for larger S it’s not so easy to visualize…
Example 2.1.9 (Graphs are metric spaces) Any connected simple graph G can be
made into a metric space by defining the distance between two vertices to be the
graph-theoretic distance between them. (The discrete metric is the special case when G
is the complete graph on S.)
Question 2.1.10.Check the conditions of a metric space for the metrics on the discrete spaceand for the connected graph.
Abuse of Notation 2.1.11.From now on, we will refer to ℝn with the Euclidean
metric by just ℝn. Moreover, if we wish to take the metric space for a subset S ⊆ ℝn
with the inherited metric, we will just write S.
2.2Convergence in metric spaces
Prototypical example for this section:The sequence(for n = 1,2,…) in ℝ.
Since we can talk about the distance between two points, we can talk about what it means
for a sequence of points to converge. This is the same as the typical epsilon-delta definition,
with absolute values replaced by the distance function.
Definition 2.2.1.Let (xn)n≥1 be a sequence of points in a metric space M. We say
that xnconverges to x if the following condition holds: for all 𝜀 > 0, there is an integer
N (depending on 𝜀) such that d(xn,x) < 𝜀 for each n ≥ N. This is written
or more verbosely as
We say that a sequence converges in M if it converges to a point in M.
You should check that this definition coincides with your intuitive notion of “converges”.
Abuse of Notation 2.2.2.If the parent space M is understood, we will allow ourselves
to abbreviate “converges in M” to just “converges”. However, keep in mind that
convergence is defined relative to the parent space; the “limit” of the space must actually
be a point in M for a sequence to converge.
Example 2.2.3 Consider the sequence x1 = 1, x2 = 1.4, x3 = 1.41, x4 = 1.414,
….
(a)
If we view this as a sequence in ℝ, it converges to .
(b)
However, even though each xi is in ℚ, this sequence does NOT converge when we
view it as a sequence in ℚ!
Question 2.2.4.What are the convergent sequences in a discrete metric space?
2.3Continuous maps
In calculus you were also told (or have at least heard) of what it means for a function to be
continuous. Probably something like
A function f : ℝ → ℝ is continuous at a point p ∈ ℝ if for every 𝜀 > 0 there
exists a δ > 0 such that < δ< 𝜀.
Question 2.3.1.Can you guess what the corresponding definition for metric spaces is?
All we have do is replace the absolute values with the more general distance functions: this
gives us a definition of continuity for any function M → N.
Definition 2.3.2.Let M = (M,dM) and N = (N,dN) be metric spaces. A function
f : M → N is continuous at a point p ∈ M if for every 𝜀 > 0 there exists a δ > 0 such
that
Moreover, the entire function f is continuous if it is continuous at every point p ∈ M.
Notice that, just like in our definition of an isomorphism of a group, we use the metric of M
for one condition and the metric of N for the other condition.
This generalization is nice because it tells us immediately how we could carry over
continuity arguments in ℝ to more general spaces like ℂ. Nonetheless, this definition is kind of
cumbersome to work with, because it makes extensive use of the real numbers (epsilons and
deltas). Here is an equivalent condition.
Theorem 2.3.3 (Sequential continuity) A function f : M → N of metric spaces is
continuous at a point p ∈ M if and only if the following property holds: if x1, x2, …is
a sequence in M converging to p, then the sequence f(x1), f(x2), …in N converges to
f(p).
Proof.One direction is not too hard:
Exercise 2.3.4.Show that 𝜀-δ continuity implies sequential continuity at each point.
Conversely, we will prove if f is not 𝜀-δ continuous at p then it does not preserve
convergence.
If f is not continuous at p, then there is a “bad” 𝜀 > 0, which we now consider fixed. So for
each choice of δ = 1∕n, there should be some point xn which is within δ of p, but which is
mapped more than 𝜀 away from f(p). But then the sequence xn converges to p, and f(xn) is
always at least 𝜀 away from f(p), contradiction. □
Example application showcasing the niceness of sequential continuity:
Proposition 2.3.5 (Composition of continuous functions is continuous) Let f : M → N
and g: N → L be continuous maps of metric spaces. Then their composition g ∘ f is
continuous.
Proof.Dead simple with sequences: Let p ∈ M be arbitrary and let xn→ p in M. Then
f(xn) → f(p) in N and g(f(xn)) → g(f(p)) in L, QED. □
Question 2.3.6.Let M be any metric space and D a discrete space. When is a map f:D → Mcontinuous?
2.4Homeomorphisms
Prototypical example for this section:The unit circle S1is homeomorphic to the boundary ofthe unit square.
When do we consider two groups to be the same? Answer: if there’s a structure-preserving
map between them which is also a bijection. For metric spaces, we do exactly the same thing,
but replace “structure-preserving” with “continuous”.
Definition 2.4.1.Let M and N be metric spaces. A function f : M → N is a
homeomorphism if it is a bijection, and both f : M → N and its inverse f−1 : N → M
are continuous. We say M and N are homeomorphic.
Needless to say, homeomorphism is an equivalence relation.
You might be surprised that we require f−1 to also be continuous. Here’s the reason: you
can show that if ϕ is an isomorphism of groups, then ϕ−1 also preserves the group operation,
hence ϕ−1 is itself an isomorphism. The same is not true for continuous bijections, which is
why we need the new condition.
Example 2.4.2 (Homeomorphism ≠ continuous bijection)
(a)
There is a continuous bijection from [0,1) to the circle, but it has no continuous
inverse.
(b)
Let M be a discrete space with size |ℝ|. Then there is a continuous function M → ℝ
which certainly has no continuous inverse.
Note that this is the topologist’s definition of “same” – homeomorphisms are “continuous
deformations”. Here are some examples.
Example 2.4.3 (Examples of homeomorphisms)
(a)
Any space M is homeomorphic to itself through the identity map.
(b)
The old saying: a doughnut (torus) is homeomorphic to a coffee cup. (Look this up
if you haven’t heard of it.)
(c)
The unit circle S1 is homeomorphic to the boundary of the unit square. Here’s one
bijection between them, after an appropriate scaling:
Example 2.4.4 (Metrics on the unit circle) It may have seemed strange that our metric
function on S1 was the one inherited from ℝ2, meaning the distance between two points
on the circle was defined to be the length of the chord. Wouldn’t it have made more
sense to use the circumference of the smaller arc joining the two points?
In fact, it doesn’t matter: if we consider S1 with the “chord” metric and the “arc” metric,
we get two homeomorphic spaces, as the map between them is continuous.
The same goes for Sn−1 for general n.
Example 2.4.5 (Homeomorphisms really don’t preserve size) Surprisingly, the open
interval (−1,1) is homeomorphic to the real line ℝ! One bijection is given by
with the inverse being given by t arctan(t).
This might come as a surprise, since (−1,1) doesn’t look that much like ℝ; the former
is “bounded” while the latter is “unbounded”.
2.5Extended example/definition: product metric
Prototypical example for this section:ℝ × ℝ is ℝ2.
Here is an extended example which will occur later on. Let M = (M,dM) and
N = (N,dN) be metric spaces (say, M = N = ℝ). Our goal is to define a metric space on
M × N.
Let pi = (xi,yi) ∈ M × N for i = 1,2. Consider the following metrics on the set of points
M × N:
dmax(p1,p2)
: = max
dEuclid(p1,p2)
: =
dtaxicab
:= dM(x1,x2) + dN(y1,y2).
All of these are good candidates. We are about to see it doesn’t matter which one we
use:
Exercise 2.5.1.Verify that
Use this to show that the metric spaces we obtain by imposing any of the three metrics arehomeomorphic, with the homeomorphism being just the identity map.
Definition 2.5.2.Hence we will usually simply refer to the metric on M × N, called
the product metric. It will not be important which of the three metrics we select.
Example 2.5.3 (ℝ2) If M = N = ℝ, we get ℝ2, the Euclidean plane. The metric dEuclid
is the one we started with, but using either of the other two metric works fine as well.
The product metric plays well with convergence of sequences.
Proposition 2.5.4 (Convergence in the product metric is by component) We have
(xn,yn) → (x,y) if and only if xn→ x and yn→ y.
Proof.We have dmax = max and the latter
approaches zero as n →∞ if and only if dM(x,xn) → 0 and dN(y,yn) → 0. □
Let’s see an application of this:
Proposition 2.5.5 (Addition and multiplication are continuous) The addition and
multiplication maps are continuous maps ℝ × ℝ → ℝ.
Proof.For multiplication: for any n we have
xnyn
=
= xy + y(xn− x) + x(yn− y) + (xn− x)(yn− y)
≤ + + .
As n →∞, all three terms on the right-hand side tend to zero. The proof that
+: ℝ × ℝ → ℝ is continuous is similar (and easier): one notes for any n that
and both terms on the right-hand side tend to zero as n →∞. □
?? covers the other two operations, subtraction and division. The upshot of this is that,
since compositions are also continuous, most of your naturally arising real-valued functions
will automatically be continuous as well. For example, the function will be
a continuous function from ℝ → ℝ, since it can be obtained by composing +, ×,
÷.
2.6Open sets
Prototypical example for this section:The open disk x2 + y2< r2in ℝ2.
Continuity is really about what happens “locally”: how a function behaves “close to a
certain point p”. One way to capture this notion of “closeness” is to use metrics as we’ve done
above. In this way we can define an r-neighborhood of a point.
Definition 2.6.1.Let M be a metric space. For each real number r > 0 and point
p ∈ M, we define
The set Mr(p) is called an r-neighborhood of p.
We can rephrase convergence more succinctly in terms of r-neighborhoods. Specifically, a
sequence (xn) converges to x if for every r-neighborhood of x, all terms of xn eventually stay
within that r-neighborhood.
Let’s try to do the same with functions.
Question 2.6.2.In terms of r-neighborhoods, what does it mean for a function f:M → N tobe continuous at a point p ∈ M?
Essentially, we require that the pre-image of every 𝜀-neighborhood has the property that
some δ-neighborhood exists inside it. This motivates:
Definition 2.6.3.A set U ⊆ M is open in M if for each p ∈ U, some r-neighborhood
of p is contained inside U. In other words, there exists r > 0 such that Mr(p) ⊆ U.
Abuse of Notation 2.6.4.Note that a set being open is defined relative to the parent
space M. However, if M is understood we can abbreviate “open in M” to just “open”.
Figure 2.1:The set of points x2 + y2< 1 in ℝ2 is open in ℝ2.
Example 2.6.5 (Examples of open sets)
(a)
Any r-neighborhood of a point is open.
(b)
Open intervals of ℝ are open in ℝ, hence the name! This is the prototypical example
to keep in mind.
(c)
The open unit ball Bn is open in ℝn for the same reason.
(d)
In particular, the open interval (0,1) is open in ℝ. However, if we embed it in ℝ2, it
is no longer open!
(e)
The empty set ∅ and the whole set of points M are open in M.
Example 2.6.6 (Non-examples of open sets)
(a)
The closed interval [0,1] is not open in ℝ. There is no 𝜀-neighborhood of the point
0 which is contained in [0,1].
(b)
The unit circle S1 is not open in ℝ2.
Question 2.6.7.What are the open sets of the discrete space?
Here are two quite important properties of open sets.
Proposition 2.6.8 (Intersections and unions of open sets)
(a)
The intersection of finitely many open sets is open.
(b)
The union of open sets is open, even if there are infinitely many.
Question 2.6.9.Convince yourself this is true.
Exercise 2.6.10.Exhibit an infinite collection of open sets in ℝ whose intersection is the set{0}. This implies that infinite intersections of open sets are not necessarily open.
The whole upshot of this is:
Theorem 2.6.11 (Open set condition) A function f : M → N of metric spaces is
continuous if and only if the pre-image of every open set in N is open in M.
Proof.I’ll just do one direction…
Exercise 2.6.12.Show that δ-𝜀 continuity follows from the open set condition.
Now assume f is continuous. First, suppose V is an open subset of the metric space N; let
U = fpre(V ). Pick x ∈ U, so y = f(x) ∈ V ; we want an open neighborhood of x inside
U.
As V is open, there is some small 𝜀-neighborhood around y which is contained inside V . By
continuity of f, we can find a δ such that the δ-neighborhood of x gets mapped by f into the
𝜀-neighborhood in N, which in particular lives inside V . Thus the δ-neighborhood lives in U,
as desired. □
2.7Closed sets
Prototypical example for this section:The closed unit disk x2 + y2≤ r2in ℝ2.
It would be criminal for me to talk about open sets without talking about closed sets. The
name “closed” comes from the definition in a metric space.
Definition 2.7.1.Let M be a metric space. A subset S ⊆ M is closed in M if the
following property holds: let x1, x2, …be a sequence of points in S and suppose that xn
converges to x in M. Then x ∈ S as well.
Abuse of Notation 2.7.2.Same caveat: we abbreviate “closed in M” to just “closed”
if the parent space M is understood.
Here’s another way to phrase it. The limit points of a subset S ⊆ M are defined
by
Thus S is closed if and only if S = limS.
Exercise 2.7.3.Prove thatlimS is closed even if S isn’t closed. (Draw a picture.)
For this reason, limS is also called the closure of S in M, and denoted S. It is simply the
smallest closed set which contains S.
Example 2.7.4 (Examples of closed sets)
(a)
The empty set ∅ is closed in M for vacuous reasons: there are no sequences of points
with elements in ∅.
(b)
The entire space M is closed in M for tautological reasons. (Verify this!)
(c)
The closed interval [0,1] in ℝ is closed in ℝ, hence the name. Like with open sets,
this is the prototypical example of a closed set to keep in mind!
(d)
In fact, the closed interval [0,1] is even closed in ℝ2.
Example 2.7.5 (Non-examples of closed sets) Let S = (0,1) denote the open interval.
Then S is not closed in ℝ because the sequence of points
converges to 0 ∈ ℝ, but 0(0,1).
I should now warn you about a confusing part of this terminology. Firstly, “most” sets areneither open nor closed.
Example 2.7.6 (A set neither open nor closed) The half-open interval [0,1) is neither
open nor closed in ℝ.
Secondly, it’s also possible for a set to be both open and closed; this will be discussed
in ?? .
The reason for the opposing terms is the following theorem:
Theorem 2.7.7 (Closed sets are complements of open sets) Let M be a metric space, and
S ⊆ M any subset. Then the following are equivalent:
The set S is closed in M.
The complement M ∖ S is open in M.
Exercise 2.7.8 (Great).Prove this theorem! You’ll want to draw a picture to make it clearwhat’s happening: for example, you might take M = ℝ2and S to be the closed unit disk.
2.8A few harder problems to think about
Problem 2A.Let M = (M,d) be a metric space. Show that
is itself a continuous function (where M × M is equipped with the product metric).
Problem 2B.Are ℚ and ℕ homeomorphic subspaces of ℝ?
Problem 2C (Continuity of arithmetic continued). Show that subtraction is a
continuous map −: ℝ × ℝ → ℝ, and division is a continuous map ÷: ℝ × ℝ>0→ ℝ.
Problem 2D.Exhibit a function f : ℝ → ℝ such that f is continuous at x ∈ ℝ if and only if
x = 0.
Problem 2E.Prove that a function f : ℝ → ℝ which is strictly increasing
must be continuous at some point.
Part II Basic Abstract Algebra
3Homomorphisms and quotient groups
3.1Generators and group presentations
Prototypical example for this section:D2n =
Let G be a group. Recall that for some element x ∈ G, we could consider the
subgroup
of G. Here’s a more pictorial version of what we did: put x in a box, seal it tightly, andshake vigorously. Using just the element x, we get a pretty explosion that produces the
subgroup above.
What happens if we put two elements x, y in the box? Among the elements that get
produced are things like
Essentially, I can create any finite product of x, y, x−1, y−1. This leads us to define:
Definition 3.1.1.Let S be a subset of G. The subgroup generated by S, denoted ,
is the set of elements which can be written as a finite product of elements in S (and
their inverses). If = G then we say S is a set of generators for G, as the elements
of S together create all of G.
Exercise 3.1.2.Why is the condition “and their inverses” not necessary if G is a finite group?(As usual, assume Lagrange’s theorem.)
Example 3.1.3 (ℤ is the infinite cyclic group) Consider 1 as an element of ℤ = (ℤ,+).
We see = ℤ, meaning {1} generates ℤ. It’s important that −1, the inverse of 1 is
also allowed: we need it to write all integers as the sum of 1 and −1.
This gives us an idea for a way to try and express groups compactly. Why not just write
down a list of generators for the groups? For example, we could write
meaning that ℤ is just the group generated by one element.
There’s one issue: the generators usually satisfy certain properties. For example, consider
ℤ∕100ℤ. It’s also generated by a single element x, but this x has the additional property that
x100 = 1. This motivates us to write
I’m sure you can see where this is going. All we have to do is specify a set of generators and
relations between the generators, and say that two elements are equal if and only if you can
get from one to the other using relations. Such an expression is appropriately called a grouppresentation.
Example 3.1.4 (Dihedral group) The dihedral group of order 2n has a presentation
Thus each element of D2n can be written uniquely in the form rα or srα, where
α = 0,1,…,n − 1.
Example 3.1.5 (Klein four group) The Klein four group, isomorphic to ℤ∕2ℤ×ℤ∕2ℤ,
is given by the presentation
Example 3.1.6 (Free group) The free group on n elements is the group whose
presentation has n generators and no relations at all. It is denoted Fn, so
In other words, F2 = is the set of strings formed by appending finitely many copies
of a, b, a−1, b−1 together.
Question 3.1.7.Notice that F1ℤ.
Abuse of Notation 3.1.8.One might unfortunately notice that “subgroup generated
by a and b” has exactly the same notation as the free group . We’ll try to be clear
based on context which one we mean.
Presentations are nice because they provide a compact way to write down groups. They do have some
shortcomings, though.1
Example 3.1.9 (Presentations can look very different) The same group can have very
different presentations. For instance consider
(To see why this is equivalent, set x = s, y = rs.)
3.2Homomorphisms
Prototypical example for this section:The “mod out by 100” map, ℤ → ℤ∕100ℤ.
How can groups talk to each other?
Two groups are “the same” if we can write an isomorphism between them. And as we saw,
two metric spaces are “the same” if we can write a homeomorphism between them. But
what’s the group analogy of a continuous map? We simply drop the “bijection”
condition.
Definition 3.2.1.Let G = (G,⋆) and H = (H,∗) be groups. A grouphomomorphism is a map ϕ : G → H such that for any g1,g2∈ G we have
(Not to be confused with “homeomorphism” from last chapter: note the spelling.)
Example 3.2.2 (Examples of homomorphisms) Let G and H be groups.
(a)
Any isomorphism G → H is a homomorphism. In particular, the identity map
G → G is a homomorphism.
(b)
The trivial homomorphismG → H sends everything to 1H.
(c)
There is a homomorphism from ℤ to ℤ∕100ℤ by sending each integer to its residue
modulo 100.
(d)
There is a homomorphism from ℤ to itself by x10x which is injective but not
surjective.
(e)
There is a homomorphism from Sn to Sn+1 by “embedding”: every permutation on
{1,…,n} can be thought of as a permutation on {1,…,n + 1} if we simply let n + 1
be a fixed point.
(f)
A homomorphism ϕ : D12→ D6 is given by s12s6 and r12r6.
(g)
Specifying a homomorphism ℤ → G is the same as specifying just the image of the
element 1 ∈ ℤ. Why?
The last two examples illustrate something: suppose we have a presentation of G. To specify
a homomorphism G → H, we only have to specify where each generator of G goes, in such a
way that the relations are all satisfied.
Important remark: the right way to think about an isomorphism is as a “bijective
homomorphism”. To be explicit,
Exercise 3.2.3.Show that GH if and only if there exist homomorphisms ϕ:G → H andψ:H → G such that ϕ ∘ ψ =idHand ψ ∘ ϕ =idG.
So the definitions of homeomorphism of metric spaces and isomorphism of groups are not
too different.
Some obvious properties of homomorphisms follow.
Fact 3.2.4.Let ϕ: G → H be a homomorphism. Then ϕ(1G) = 1H and ϕ(g−1) =
ϕ(g)−1.
Proof.Boring, and I’m sure you could do it yourself if you wanted to. □
Now let me define a very important property of a homomorphism.
Definition 3.2.5.The kernel of a homomorphism ϕ: G → H is defined by
It is a subgroup of G (in particular, 1G∈ kerϕ for obvious reasons).
Question 3.2.6.Verify thatkerϕ is in fact a subgroup of G.
We also have the following important fact, which we also encourage the reader to
verify.
Proposition 3.2.7 (Kernel determines injectivity) The map ϕ is injective if and only if
kerϕ = {1G}.
To make this concrete, let’s compute the kernel of each of our examples.
Example 3.2.8 (Examples of kernels)
(a)
The kernel of any isomorphism G → H is trivial, since an isomorphism is injective.
In particular, the kernel of the identity map G → G is {1G}.
(b)
The kernel of the trivial homomorphism G → H (by g1H) is all of G.
(c)
The kernel of the homomorphism ℤ → ℤ∕100ℤ by nn is precisely
(d)
The kernel of the map ℤ → ℤ by x10x is trivial: {0}.
(e)
There is a homomorphism from Sn to Sn+1 by “embedding”, but it also has trivial
kernel because it is injective.
(f)
A homomorphism ϕ: D12→ D6 is given by s12s6 and r12r6. You can check
that
(g)
Exercise below.
Exercise 3.2.9.Fix any g ∈ G. Suppose we have a homomorphism ℤ → G by ngn. What isthe kernel?
Question 3.2.10.Show that for any homomorphism ϕ : G → H, the image ϕimg(G) is asubgroup of H. Hence, we’ll be especially interested in the case where ϕ is surjective.
3.3Cosets and modding out
Prototypical example for this section:Modding out by n: ℤ∕(n ⋅ ℤ)ℤ∕nℤ.
The next few sections are a bit dense. If this exposition doesn’t work for you, try[?].
Let G and Q be groups, and suppose there exists a surjective homomorphism
In other words, if ϕ is injective then ϕ : G → Q is a bijection, and hence an isomorphism.
But suppose we’re not so lucky and kerϕ is bigger than just {1G}. What is the correct
interpretation of a more general homomorphism?
Let’s look at the special case where ϕ : ℤ → ℤ∕100ℤ is “modding out by 100”. We already
saw that the kernel of this map is
Recall now that kerϕ is a subgroup of G. What this means is that ϕ is indifferent to thesubgroup 100ℤ of ℤ:
So ℤ∕100ℤ is what we get when we “mod out by 100”. Cool.
In other words, let G be a group and ϕ : G ↠ Q be a surjective homomorphism with kernel
N ⊆ G.
We claim that Q should be thought of as the quotient of G by N.
To formalize this, we will define a so-called quotient groupG∕N in terms of G and N only
(without referencing Q) which will be naturally isomorphic to Q.
For motivation, let’s give a concrete description of Q using just ϕ and G. Continuing our
previous example, let N = 100ℤ be our subgroup of G. Consider the sets
N
=
1 + N
=
2 + N
=
99 + N
= .
The elements of each set all have the same image when we apply ϕ, and moreover any two
elements in different sets have different images. Then the main idea is to notice
that
We can think of Q as the group whose elements are the sets above.
Thus, given ϕ we define an equivalence relation ∼N on G by saying x ∼Ny for ϕ(x) = ϕ(y).
This ∼N divides G into several equivalence classes in G which are in obvious bijection
with Q, as above. Now we claim that we can write these equivalence classes very
explicitly.
Exercise 3.3.1.Show that x ∼Ny if and only if x = yn for some n ∈ N (in the mod 100example, this means they “differ by some multiple of 100”). Thus for any g ∈ G, the equivalenceclass of ∼Nwhich contains g is given explicitly by
Here’s the word that describes the types of sets we’re running into now.
Definition 3.3.2.Let H be any subgroup of G (not necessarily the kernel of some
homomorphism). A set of the form gH is called a left coset of H.
Remark 3.3.3 — Although the notation might not suggest it, keep in mind that g1N
is often equal to g2N even if g1≠g2. In the “mod 100” example, 3 + N = 103 + N. In
other words, these cosets are sets.
This means that if I write “let gH be a coset” without telling you what g is, you can’t
figure out which g I chose from just the coset itself. If you don’t believe me, here’s an
example of what I mean:
There’s no reason to think I picked x = 3. (I actually picked x = −13597.)
Remark 3.3.4 — Given cosets g1H and g2H, you can check that the map xg2g1−1x
is a bijection between them. So actually, all cosets have the same cardinality.
So, long story short,
Elements of the group Q are naturally identified with left cosets of N.
In practice, people often still prefer to picture elements of Q as single points (for example
it’s easier to think of ℤ∕2ℤ as {0,1} rather than {…,−2,0,2,…},{…,−1,1,3,…}). If you like
this picture, then you might then draw G as a bunch of equally tall fibers (the cosets), which
are then “collapsed” onto Q.
Now that we’ve done this, we can give an intrinsic definition for the quotient group we
alluded to earlier.
Definition 3.3.5.A subgroup N of G is called normal if it is the kernel of some
homomorphism. We write this as N ⊴ G.
Definition 3.3.6.Let N ⊴ G. Then the quotient group, denoted G∕N (and read “G mod
N”), is the group defined as follows.
The elements of G∕N will be the left cosets of N.
We want to define the product of two cosets C1 and C2 in G∕N. Recall that the
cosets are in bijection with elements of Q. So let q1 be the value associated to the
coset C1, and q2 the one for C2. Then we can take the product to be the coset
corresponding to q1q2.
Quite importantly, we can also do this in terms of representatives of thecosets. Let g1∈ C1 and g2∈ C2, so C1 = g1N and C2 = g2N. Then C1⋅ C2
should be the coset which contains g1g2. This is the same as the above definition
since ϕ(g1g2) = ϕ(g1)ϕ(g2) = q1q2; all we’ve done is define the product in terms
of elements of G, rather than values in H.
Using the gN notation, and with ?? in mind, we can write this even more
succinctly:
And now you know why the integers modulo n are often written ℤ∕nℤ!
Question 3.3.7.Take a moment to digest the above definition.
By the way we’ve built it, the resulting group G∕N is isomorphic to Q. In a sense we think
of G∕N as “G modulo the condition that n = 1 for all n ∈ N”.
3.4(Optional) Proof of Lagrange’s theorem
As an aside, with the language of cosets we can now show Lagrange’s theorem in the general
case.
Theorem 3.4.1 (Lagrange’s theorem) Let G be a finite group, and let H be any subgroup.
Then divides .
The proof is very simple: note that the cosets of H all have the same size and form a
partition of G (even when H is not necessarily normal). Hence if n is the number of cosets,
then n ⋅ = .
Question 3.4.2.Conclude that x= 1 by taking H = ⊆ G.
Remark 3.4.3 — It should be mentioned at this point that in general, if G is a finite
group and N is normal, then |G∕N| = |G|∕|N|.
3.5Eliminating the homomorphism
Prototypical example for this section:Again ℤ∕nℤℤ∕nℤ.
Let’s look at the last definition of G∕N we provided. The short version is:
The elements of G∕N are cosets gN, which you can think of as equivalence classes
of a relation ∼N (where g1∼Ng2 if g1 = g2n for some n ∈ N).
Given cosets g1N and g2N the group operation is
Question: where do we actually use the fact that N is normal? We don’t talk about ϕ or Q
anywhere in this definition.
The answer is in ?? . The group operation takes in two cosets, so it doesn’t know what g1
and g2 are. But behind the scenes, the normal condition guarantees that the groupoperation can pick any g1and g2it wants and still end up with the same coset. If
we didn’t have this property, then it would be hard to define the product of two
cosets C1 and C2 because it might make a difference which g1∈ C1 and g2∈ C2
we picked. The fact that N came from a homomorphism meant we could pick any
representatives g1 and g2 of the cosets we wanted, because they all had the same
ϕ-value.
We want some conditions which force this to be true without referencing ϕ at all. Suppose
ϕ: G → K is a homomorphism of groups with H = kerϕ. Aside from the fact H is a group, we
can get an “obvious” property:
Question 3.5.1.Show that if h ∈ H, g ∈ G, then ghg−1∈ H. (Check ϕ(ghg−1) = 1K.)
Example 3.5.2 (Example of a non-normal subgroup) Let
D12 = . Consider the subgroup of order two H = {1,s}
and notice that
Hence H is not normal, and cannot be the kernel of any homomorphism.
Well, duh – so what? Amazingly it turns out that that this is the sufficient condition we
want. Specifically, it makes the nice “coset multiplication” we wanted work out.
Remark 3.5.3 (For math contest enthusiasts) — This coincidence is really a lot like
functional equations at the IMO. We all know that normal subgroups H satisfy
ghg−1∈ H; the surprise is that from the latter seemingly weaker condition, we can
deduce H is normal.
Thus we have a new criterion for “normal” subgroups which does not make any external
references to ϕ.
Theorem 3.5.4 (Algebraic condition for normal subgroups) Let H be a subgroup of G. Then the
following are equivalent:
H ⊴ G.
For every g ∈ G and h ∈ H, ghg−1∈ H.
Proof.We already showed one direction.
For the other direction, we need to build a homomorphism with kernel H. So we simply
define the group G∕H as the cosets. To put a group operation, we need to verify:
Claim 3.5.5.If g1′∼Hg1 and g2′∼Hg2 then g1′g2′∼Hg1g2.
Proof. Boring algebraic manipulation (again functional equation style). Let g1′ = g1h1
and g2′ = g2h2, so we want to show that g1h1g2h2∼Hg1g2. Since H has the property,
g2−1h1g2 is some element of H, say h3. Thus h1g2 = g2h3, and the left-hand side becomes
g1g2(h3h2), which is fine since h3h2∈ H. ■
With that settled we can just define the product of two cosets (of normal subgroups)
by
Thus the claim above shows that this multiplication is well-defined (this verification is the
“content” of the theorem). So G∕H is indeed a group! Moreover there is an obvious
“projection” homomorphism G → G∕H (with kernel H), by ggH. □
Example 3.5.6 (Modding out in the product group) Consider again the product group
G × H. Earlier we identified a subgroup
You can easily see that G′⊴ G × H. (Easy calculation.)
Moreover, just as the notation would imply, you can check that
Indeed, we have (g,h) ∼G′(1G,h) for all g ∈ G and h ∈ H.
Example 3.5.7 (Another explicit computation) Let ϕ : D8→ ℤ∕4ℤ be defined by
The kernel of this map is N = {1,r2,sr,sr3}.
We can do a quick computation of all the elements of D8 to get
The two relevant fibers are
and
So we see that |D8∕N| = 2 is a group of order two, or ℤ∕2ℤ. Indeed, the image of ϕ is
Question 3.5.8.Suppose G is abelian. Why does it follow that any subgroup of G is normal?
Finally here’s some food for thought: suppose one has a group presentation for a group G
that uses n generators. Can you write it as a quotient of the form Fn∕N, where N is a normal
subgroup of Fn?
3.6(Digression) The first isomorphism theorem
One quick word about what other sources usually say.
Most textbooks actually define normal using the ghg−1∈ H property. Then they define
G∕H for normal H in the way I did above, using the coset definition
Using purely algebraic manipulations (like I did) this is well-defined, and so now you have
this group G∕H or something. The underlying homomorphism isn’t mentioned at all, or is just
mentioned in passing.
I think this is incredibly dumb. The normal condition looks like it gets pulled out of thin air
and no one has any clue what’s going on, because no one has any clue what a normal subgroup
actually should look like.
Other sources like to also write the so-called first isomorphism
theorem.2
It goes like this.
Theorem 3.6.1 (First isomorphism theorem) Let ϕ : G → H be a homomorphism. Then
G∕kerϕ is isomorphic to ϕimg(G).
To me, this is just a clumsier way of stating the same idea.
About the only merit this claim has is that if ϕ is injective, then the image ϕimg(G) is an
isomorphic copy of G inside the group H. (Try to see this directly!) This is a pattern we’ll
often see in other branches of mathematics: whenever we have an injective structure-preservingmap, often the image of this map will be some “copy” of G. (Here “structure” refers to the
group multiplication, but we’ll see some more other examples of “types of objects”
later!)
In that sense an injective homomorphism ϕ : GH is an embedding of G into
H.
3.7A few harder problems to think about
Problem 3A (18.701 at MIT).Determine all groups G for which the map ϕ : G → G
defined by
is a homomorphism.
Problem 3B.Consider the dihedral group G = D10.
(a)
Is H = a normal subgroup of G? If so, compute G∕H up to isomorphism.
(b)
Is H = a normal subgroup of G? If so, compute G∕H up to isomorphism.
Problem 3C.Does S4 have a normal subgroup of order 3?
Problem 3D.Let G and H be finite groups, where = 1000 and = 999. Show that a
homomorphism G → H must be trivial.
Problem 3E.Let ℂ× denote the nonzero complex numbers under multiplication.
Show that there are five homomorphisms ℤ∕5ℤ → ℂ× but only two homomorphisms
D10→ ℂ×, even though ℤ∕5ℤ is a subgroup of D10.
Problem 3F.Find a non-abelian group G such that every subgroup of G is
normal. (These groups are called Hamiltonian.)
Problem 3G (PRIMES entrance exam, 2017).Let G be a group
with presentation given by
Determine the order of G.
Problem 3H (Homophony group).The homophony group (of English) is
the group with 26 generators a, b, …, z and one relation for every pair of English words which
sound the same. For example knight = night (and hence k = 1). Prove that the group is
trivial.
4Rings and ideals
4.1Some motivational metaphors about rings vs groups
In this chapter we’ll introduce the notion of a commutative ring R. It is a larger structure
than a group: it will have two operations addition and multiplication, rather than just one.
We will then immediately define a ring homomorphism R → S between pairs of
rings.
This time, instead of having normal subgroups H ⊴ G, rings will instead have subsets I ⊆ R
called ideals, which are not themselves rings but satisfy some niceness conditions. We will
then show how you to define R∕I, in analogy to G∕H as before. Finally, like with groups, we
will talk a bit about how to generate ideals.
Here is a possibly helpful table of analogies to help you keep track:
Group
Ring
Notation
G
R
Operations
⋅
+, ×
Commutativity
only if abelian
for us, always
Sub-structure
subgroup
(not discussed)
Homomorphism
grp hom. G → H
ring hom. R → S
Kernel
normal subgroup
ideal
Quotient
G∕H
R∕I
4.2(Optional) Pedagogical notes on motivation
I wrote most of these examples with a number theoretic eye in mind; thus if you liked
elementary number theory, a lot of your intuition will carry over. Basically, we’ll try to
generalize properties of the ring ℤ to any abelian structure in which we can also multiply.
That’s why, for example, you can talk about “irreducible polynomials in ℚ[x]” in
the same way you can talk about “primes in ℤ”, or about “factoring polynomials
modulo p” in the same way we can talk “unique factorization in ℤ”. Even if you only
care about ℤ (say, you’re a number theorist), this has a lot of value: I assure you
that trying to solve xn + yn = zn (for n > 2) requires going into a ring other than
ℤ!
Thus for all the sections that follow, keep ℤ in mind as your prototype.
I mention this here because commutative algebra is also closely tied to algebraic geometry.
Lots of the ideas in commutative algebra have nice “geometric” interpretations that motivate
the definitions, and these connections are explored in the corresponding part later. So, I want
to admit outright that this is not the only good way (perhaps not even the most natural one)
of motivating what is to follow.
4.3Definition and examples of rings
Prototypical example for this section:ℤ all the way! Also R[x] and various fields (nextsection).
Definition 4.3.1.A ring is a triple (R,+,×), the two operations usually called addition and
multiplication, such that
(i)
(R,+) is an abelian group, with identity 0R, or just 0.
(ii)
× is an associative, binary operation on R with some identity, written 1R or just
1.
(iii)
Multiplication distributes over addition.
The ring R is commutative if × is commutative.
Abuse of Notation 4.3.2.As usual, we will abbreviate (R,+,×) to R.
Abuse of Notation 4.3.3.For simplicity, assume all rings are commutative for the
rest of this chapter. We’ll run into some noncommutative rings eventually, but for such
rings we won’t need the full theory of this chapter anyways.
These definitions are just here for completeness. The examples are much more
important.
Example 4.3.4 (Typical rings)
(a)
The sets ℤ, ℚ, ℝ and ℂ are all rings with the usual addition and multiplication.
(b)
The integers modulo n are also a ring with the usual addition and multiplication.
We also denote it by ℤ∕nℤ.
Here is also a trivial example.
Definition 4.3.5.The zero ring is the ring R with a single element. We denote the
zero ring by 0. A ring is nontrivial if it is not the zero ring.
Exercise 4.3.6 (Comedic).Show that a ring is nontrivial if and only if 0R≠1R.
Since I’ve defined this structure, I may as well state the obligatory facts about it.
Fact 4.3.7.For any ring R and r ∈ R, r ⋅ 0R = 0R. Moreover, r ⋅ (−1R) = −r.
Here are some more examples of rings.
Example 4.3.8 (Product ring) Given two rings R and S the product ring, denoted
R × S, is defined as ordered pairs (r,s) with both operations done component-wise. For
example, the Chinese remainder theorem says that
with the isomorphism n mod 15(n mod 3,n mod 5).
Remark 4.3.9 — Equivalently, we can define R × S as the abelian group R ⊕ S, and
endow it with the multiplication where r ⋅ s = 0 for r ∈ R, s ∈ S.
Question 4.3.10.Which (r,s) is the identity element of the product ring R × S?
Example 4.3.11 (Polynomial ring) Given any ring R, the polynomial ringR[x] is
defined as the set of polynomials with coefficients in R:
This is pronounced “R adjoin x”. Addition and multiplication are done exactly in the
way you would expect.
Remark 4.3.12 (Digression on division) — Happily, polynomial division also does what
we expect: if p ∈ R[x] is a polynomial, and p(a) = 0, then (x − a)q(x) = p(x) for some
polynomial q. Proof: do polynomial long division.
With that, note the caveat that
has four roots 1, 3, 5, 7 in ℤ∕8ℤ.
The problem is that 2⋅4 = 0 even though 2 and 4 are not zero; we call 2 and 4 zero divisors
for that reason. In an integral domain (a ring without zero divisors), this pathology goes
away, and just about everything you know about polynomials carries over. (I’ll say this
all again next section.)
Example 4.3.13 (Multi-variable polynomial ring) We can consider polynomials in n
variables with coefficients in R, denoted R[x1,…,xn]. (We can even adjoin infinitely many
x’s if we like!)
Example 4.3.14 (Gaussian integers are a ring) The Gaussian integers are the set of
complex numbers with integer real and imaginary parts, that is
Abuse of Notation 4.3.15 (Liberal use of adjoinment).Careful readers will detect
some abuse in notation here. ℤ[i] should officially be “integer-coefficient polynomials in
a variable i”. However, it is understood from context that i2 = −1; and a polynomial in
i = “is” a Gaussian integer.
Example 4.3.16 (Cube root of 2) As another example (using the same abuse of
notation):
4.4Fields
Prototypical example for this section:ℚ is a field, but ℤ is not.
Although we won’t need to know what a field is until next chapter, they’re so convenient for
examples I will go ahead and introduce them now.
As you might already know, if the multiplication is invertible, then we call the ring a field.
To be explicit, let me write the relevant definitions.
Definition 4.4.1. A unit of a ring R is an element u ∈ R which is invertible: for some
x ∈ R we have ux = 1R.
Example 4.4.2 (Examples of units)
(a)
The units of ℤ are ±1, because these are the only things which “divide 1” (which is
the reason for the name “unit”).
(b)
On the other hand, in ℚ everything is a unit (except 0). For example, is a unit
since ⋅ = 1.
(c)
The Gaussian integers ℤ[i] have four units: ±1 and ±i.
Definition 4.4.3.A nontrivial (commutative) ring is a field when all its nonzero
elements are units.
Colloquially, we say that
A field is a structure where you can add, subtract, multiply, and divide.
Depending on context, they are often denoted either k, K, F.
Example 4.4.4 (First examples of fields)
(a)
ℚ, ℝ, ℂ are fields, since the notion makes sense in them.
(b)
If p is a prime, then ℤ∕pℤ is a field, which we denote will usually denote by 𝔽p.
The trivial ring 0 is not considered a field, since we require fields to be nontrivial.
4.5Homomorphisms
Prototypical example for this section:ℤ → ℤ∕5ℤ by modding out by 5.
This section is going to go briskly – it’s the obvious generalization of all the stuff we did with quotient
groups.2
First, we define a homomorphism and isomorphism.
Definition 4.5.1.Let R = (R,+R,×R) and S = (S,+S,×S) be rings. A ring homomorphism
is a map ϕ : R → S such that
(i)
ϕ(x + Ry) = ϕ(x) + Sϕ(y) for each x,y ∈ R.
(ii)
ϕ(x ×Ry) = ϕ(x) ×Sϕ(y) for each x,y ∈ R.
(iii)
ϕ(1R) = 1S.
If ϕ is a bijection then ϕ is an isomorphism and we say that rings R and S are isomorphic.
Just what you would expect. The only surprise is that we also demand ϕ(1R) to go to
1S. This condition is not extraneous: consider the map ℤ → ℤ called “multiply by
zero”.
Example 4.5.2 (Examples of homomorphisms)
(a)
The identity map, as always.
(b)
The map ℤ → ℤ∕5ℤ modding out by 5.
(c)
The map ℝ[x] → ℝ by p(x)p(0) by taking the constant term.
(d)
For any ring R, there is a trivial ring homomorphism R → 0.
Example 4.5.3 (Non-examples of homomorphisms) Because we require 1R to 1S, some maps
that you might have thought were homomorphisms will fail.
(a)
The map ℤ → ℤ by x2x is not a ring homomomorphism. Aside from the fact it
sends 1 to 2, it also does not preserve multiplication.
(b)
If S is a nontrivial ring, the map R → S by x0 is not a ring homomorphism, even
though it preserves multiplication.
(c)
There is no ring homomorphism ℤ∕2016ℤ → ℤ at all.
In particular, whereas for groups G and H there was always a trivial group homomorphism
sending everything in G to 1H, this is not the case for rings.
4.6Ideals
Prototypical example for this section:The multiples of 5 are an ideal of ℤ.
Now, just like we were able to mod out by groups, we’d also like to define quotient rings. So
once again,
Definition 4.6.1.The kernel of a ring homomorphism ϕ: R → S, denoted kerϕ, is
the set of r ∈ R such that ϕ(r) = 0.
In group theory, we were able to characterize the “normal” subgroups by a few obviously
necessary conditions (namely, gHg−1 = H). We can do the same thing for rings, and it’s in
fact easier because our operations are commutative.
First, note two obvious facts:
If ϕ(x) = ϕ(y) = 0, then ϕ(x + y) = 0 as well. So kerϕ should be closed under
addition.
If ϕ(x) = 0, then for any r ∈ R we have ϕ(rx) = ϕ(r)ϕ(x) = 0 too. So for x ∈ kerϕ
and any r ∈ R, we have rx ∈ kerϕ.
A (nonempty) subset I ⊆ R is called an ideal if it satisfies these properties. That is,
Definition 4.6.2.A nonempty subset I ⊆ R is an ideal if it is closed under addition,
and for each x ∈ I, rx ∈ I for all r ∈ R. It is proper if I≠R.
Note that in the second condition, r need not be in I! So this is stronger than merely saying
I is closed under multiplication.
Remark 4.6.3 — If R is not commutative, we also need the condition xr ∈ I. That is,
the ideal is two-sided: it absorbs multiplication from both the left and the right. But
since rings in Napkin are commutative we needn’t worry with this distinction.
Example 4.6.4 (Prototypical example of an ideal) Consider the set I = 5ℤ =
{…,−10,−5,0,5,10,…} as an ideal in ℤ. We indeed see I is the kernel of the “take mod
5” homomorphism:
It’s clearly closed under addition, but it absorbs multiplication from all elements of ℤ:
given 15 ∈ I, 999 ∈ ℤ, we get 15 ⋅ 999 ∈ I.
Exercise 4.6.5 (Mandatory: fields have two ideals).If K is a field, show that K has exactly twoideals. What are they?
Now we claim that these conditions are sufficient. More explicitly,
Theorem 4.6.6 (Ring analog of normal subgroups) Let R be a ring and I ⊊ R. Then I is
the kernel of some homomorphism if and only if it’s an ideal.
Proof.It’s quite similar to the proof for the normal subgroup thing, and you might try
it yourself as an exercise.
Obviously the conditions are necessary. To see they’re sufficient, we define a ring by
“cosets”
These are the equivalence where we say r1∼ r2 if r1−r2∈ I (think of this as taking “mod
I”). To see that these form a ring, we have to check that the addition and multiplication we
put on them is well-defined. Specifically, we want to check that if r1∼ s1 and r2∼ s2, then
r1 + r2∼ s1 + s2 and r1r2∼ s1s2. We actually already did the first part – just think of R and
S as abelian groups, forgetting for the moment that we can multiply. The multiplication is
more interesting.
Exercise 4.6.7 (Recommended).Show that if r1∼ s1and r2∼ s2, then r1r2∼ s1s2. You willneed to use the fact that I absorbs multiplication from any elements of R, not just those in I.
Anyways, since this addition and multiplication is well-defined there is now a surjective
homomorphism R → S with kernel exactly I. □
Definition 4.6.8.Given an ideal I, we define as above the quotient ring
It’s the ring of these equivalence classes. This ring is pronounced “R mod I”.
Example 4.6.9 (ℤ∕5ℤ) The integers modulo 5 formed by “modding out additively by
5” are the ℤ∕5ℤ we have already met.
But here’s an important point: just as we don’t actually think of ℤ∕5ℤ as consisting of
k + 5ℤ for k = 0,…,4, we also don’t really want to think about R∕I as elements r + I. The
better way to think about it is
R∕I is the result when we declare that elements of I are all zero; that is, we “modout by elements of I”.
For example, modding out by 5ℤ means that we consider all elements in ℤ divisible by 5 to
be zero. This gives you the usual modular arithmetic!
Exercise 4.6.10.Earlier, we wrote ℤ[i] for the Gaussian integers, which was a slight abuse ofnotation. Convince yourself that this ring could instead be written as ℤ[x]∕(x2+1), if we wantedto be perfectly formal. (We will stick with ℤ[i] though — it’s more natural.)
Figure out the analogous formalization of ℤ[].
4.7Generating ideals
Prototypical example for this section:In ℤ, the ideals are all of the form (n).
Let’s give you some practice with ideals.
An important piece of intuition is that once an ideal contains a unit, it contains 1, and thus
must contain the entire ring. That’s why the notion of “proper ideal” is useful language. To
expand on that:
Proposition 4.7.1 (Proper ideal⇔no units) Let R be a ring and I ⊆ R an ideal. Then I
is proper (i.e. I≠R) if and only if it contains no units of R.
Proof.Suppose I contains a unit u, i.e. an element u with an inverse u−1. Then it
contains u ⋅ u−1 = 1, and thus I = R. Conversely, if I contains no units, it is obviously
proper. □
As a consequence, if K is a field, then its only ideals are (0) and K (this was ?? ). So for
our practice purposes, we’ll be working with rings that aren’t fields.
First practice: ℤ.
Exercise 4.7.2.Show that the only ideals of ℤ are precisely those sets of the form nℤ, where nis a nonnegative integer.
Thus, while ideals of fields are not terribly interesting, ideals of ℤ look eerily like elements of
ℤ. Let’s make this more precise.
Definition 4.7.3.Let R be a ring. The ideal generated by a set of elements x1,…,xn∈R is denoted by I = (x1,x2,…,xn) and given by
One can think of this as “the smallest ideal containing all the xi”.
The analogy of putting the {xi} in a sealed box and shaking vigorously kind of works here
too.
Remark 4.7.4 (Linear algebra digression) — If you know linear algebra, you can
summarize this as: an ideal is an R-module. The ideal (x1,…,xn) is the submodule
spanned by x1,…,xn.
In particular, if I = (x) then I consists of exactly the “multiples of x”, i.e. numbers of the
form rx for r ∈ R.
Remark 4.7.5 — We can also apply this definition to infinite generating sets, as long
as only finitely many of the ri are not zero (since infinite sums don’t make sense in
general).
Example 4.7.6 (Examples of generated ideals)
(a)
As (n) = nℤ for all n ∈ ℤ, every ideal in ℤ is of the form (n).
(b)
In ℤ[i], we have (5) = .
(c)
In ℤ[x], the ideal (x) consists of polynomials with zero constant terms.
(d)
In ℤ[x,y], the ideal (x,y) again consists of polynomials with zero constant terms.
(e)
In ℤ[x], the ideal (x,5) consists of polynomials whose constant term is divisible by
5.
Question 4.7.7.Please check that the set I = is indeed always anideal (closed under addition, and absorbs multiplication).
Now suppose I = (x1,…,xn). What does R∕I look like? According to what I said at the end
of the last section, it’s what happens when we “mod out” by each of the elements xi. For
example…
Example 4.7.8 (Modding out by generated ideals)
(a)
Let R = ℤ and I = (5). Then R∕I is literally ℤ∕5ℤ, or the “integers modulo 5”: it
is the result of declaring 5 = 0.
(b)
Let R = ℤ[x] and I = (x). Then R∕I means we send x to zero; hence R∕Iℤ as
given any polynomial p(x) ∈ R, we simply get its constant term.
(c)
Let R = ℤ[x] again and now let I = (x − 3). Then R∕I should be thought of as the
quotient when x − 3 ≡ 0, that is, x ≡ 3. So given a polynomial p(x) its image after
we mod out should be thought of as p(3). Again R∕Iℤ, but in a different way.
(d)
Finally, let I = (x − 3,5). Then R∕I not only sends x to three, but also 5 to zero.
So given p ∈ R, we get p(3) (mod 5). Then R∕Iℤ∕5ℤ.
Remark 4.7.9 (Mod notation) — By the way, given an ideal I of a ring R, it’s totally
legit to write
to mean that x − y ∈ I. Everything you learned about modular arithmetic carries over.
4.8Principal ideal domains
Prototypical example for this section:ℤ is a PID, ℤ[x] is not. ℂ[x] is a PID, ℂ[x,y] isnot.
What happens if we put multiple generators in an ideal, like (10,15) ⊆ ℤ? Well, we have by
definition that (10,15) is given as a set by
If you’re good at number theory you’ll instantly recognize this as 5ℤ = (5). Surprise! In ℤ,
the ideal (a,b) is exactly gcd(a,b)ℤ. And that’s exactly the reason you often see the GCD of
two numbers denoted (a,b).
We call such an ideal (one generated by a single element) a principal ideal. So, in ℤ, every
ideal is principal. But the same is not true in more general rings.
Example 4.8.1 (A non-principal ideal) In ℤ[x], I = (x,2015) is not a principal ideal.
For if I = (f) for some polynomial f ∈ I then f divides x and 2015. This can only occur
if f = ±1, but then I contains ±1, which it does not.
A ring with the property that all its ideals are principal is called a principal ideal ring.
We like this property because they effectively let us take the “greatest common factor” in a
similar way as the GCD in ℤ.
In practice, we actually usually care about so-called principal ideal domains (PID’s).
But we haven’t defined what a domain is yet. Nonetheless, all the examples below are actually
PID’s, so we will go ahead and use this word for now, and tell you what the additional
condition is in the next chapter.
Example 4.8.2 (Examples of PID’s) To reiterate, for now you should just verify that these
are principal ideal rings, even though we are using the word PID.
(a)
As we saw, ℤ is a PID.
(b)
As we also saw, ℤ[x] is not a PID, since I = (x,2015) for example is not principal.
(c)
It turns out that for a field k the ring k[x] is always a PID. For example, ℚ[x], ℝ[x],
ℂ[x] are PID’s.
If you want to try and prove this, first prove an analog of Bezout’s lemma, which
implies the result.
(d)
ℂ[x,y] is not a PID, because (x,y) is not principal.
4.9Noetherian rings
Prototypical example for this section:ℤ[x1,x2,…] is not Noetherian, but most reasonable ringsare. In particular polynomial rings are. (Equivalently, only weirdos care about non-Noetherianrings).
If it’s too much to ask that an ideal is generated by one element, perhaps we can at least
ask that our ideals are generated by finitely many elements. Unfortunately, in certain weird
rings this is also not the case.
Example 4.9.1 (Non-Noetherian ring) Consider the ring R = ℤ[x1,x2,x3,…] which has
infinitely many free variables. Then the ideal I = (x1,x2,…) ⊆ R cannot be written with
a finite generating set.
Nonetheless, most “sane” rings we work in do have the property that their ideals are finitely
generated. We now name such rings and give two equivalent definitions:
Proposition 4.9.2 (The equvialent definitions of a Noetherian ring) For a ring R, the following
are equivalent:
(a)
Every ideal I of R is finitely generated (i.e. can be written with a finite generating
set).
(b)
There does not exist an infinite ascending chain of ideals
The absence of such chains is often called the ascending chain condition.
Such rings are called Noetherian.
Example 4.9.3 (Non-Noetherian ring breaks ACC) In the ring R = ℤ[x1,x2,x3,…] we
have an infinite ascending chain
From the example, you can kind of see why the proposition is true: from an
infinitely generated ideal you can extract an ascending chain by throwing
elements in one at a time. I’ll leave the proof to you if you want to do
it.3
Question 4.9.4.Why are fields Noetherian? Why are PID’s (such as ℤ) Noetherian?
This leaves the question: is our prototypical non-example of a PID, ℤ[x], a Noetherian ring?
The answer is a glorious yes, according to the celebrated Hilbert basis theorem.
Theorem 4.9.5 (Hilbert basis theorem) Given a Noetherian ring R, the ring R[x] is also
Noetherian. Thus by induction, R[x1,x2,…,xn] is Noetherian for any integer n.
The proof of this theorem is really olympiad flavored, so I couldn’t possibly spoil it – I’ve
left it as a problem at the end of this chapter.
Noetherian rings really shine in algebraic geometry, and it’s a bit hard for me to motivate
them right now, other than to say “most rings you’ll encounter are Noetherian”. Please bear
with me!
4.10A few harder problems to think about
Problem 4A.The ring R = ℝ[x]∕(x2 + 1) is one that you’ve seen before. What is its name?
Problem 4B.Show that ℂ[x]∕(x2− x)ℂ × ℂ.
Problem 4C.In the ring ℤ, let I = (2016) and J = (30). Show that I ∩ J is an ideal
of ℤ and compute its elements.
Problem 4D⋆. Let R be a ring and I an ideal. Find an inclusion-preserving bijection
between
ideals of R∕I, and
ideals of R which contain I.
Problem 4E.Let R be a ring.
(a)
Prove that there is exactly one ring homomorphism ℤ → R.
(b)
Prove that the number of ring homomorphisms ℤ[x] → R is equal to the number of
elements of R.
Problem 4F.Prove the Hilbert basis theorem, ?? .
Problem 4G (USA Team Selection Test 2016).Let 𝔽p denote the integers modulo a
fixed prime number p. Define Ψ: 𝔽p[x] → 𝔽p[x] by
Let S denote the image of Ψ.
(a)
Show that S is a ring with addition given by polynomial addition, and multiplication
given by function composition.
(b)
Prove that Ψ: 𝔽p[x] → S is then a ring isomorphism.
Problem 4H.Let A ⊆ B ⊆ C be rings. Suppose C is a finitely generated
A-module. Does it follow that B is a finitely generated A-module?
5Flavors of rings
We continue our exploration of rings by considering some nice-ness properties that rings or
ideals can satisfy, which will be valuable later on. As before, number theory is interlaced as
motivation. I guess I can tell you at the outset what the completed table is going to look like,
so you know what to expect.
Ring noun
Ideal adjective
Relation
PID
principal
R is a PID ⇔R is an integral domain,
and every I is principal
Noetherian ring
finitely generated
R is Noetherian ⇔ every I is fin. gen.
field
maximal
R∕I is a field ⇔I is maximal
integral domain
prime
R∕I is an integral domain ⇔I is prime
5.1Fields
Prototypical example for this section:ℚ is a field, but ℤ is not.
We already saw this definition last chapter: a field K is a nontrivial ring for which every
nonzero element is a unit.
In particular, there are only two ideals in a field: the ideal (0), which is maximal, and the
entire field K.
5.2Integral domains
Prototypical example for this section:ℤ is an integral domain.
In practice, we are often not so lucky that we have a full-fledged field. Now it would be nice
if we could still conclude the zero product property: if ab = 0 then either a = 0 or b = 0. If our
ring is a field, this is true: if b≠0, then we can multiply by b−1 to get a = 0. But many
other rings we consider like ℤ and ℤ[x] also have this property, despite not having
division.
Not all rings though: in ℤ∕15ℤ,
If a,b≠0 but ab = 0 then we say a and b are zero divisors of the ring R. So we give a name
to such rings.
Definition 5.2.1.A nontrivial ring with no zero divisors is called an integral domain.1
Question 5.2.2.Show that a field is an integral domain.
Exercise 5.2.3 (Cancellation in integral domains).Suppose ac = bc in an integral domain, andc≠0. Show that that a = b. (There is no c−1to multiply by, so you have to use the definition.)
Example 5.2.4 (Examples of integral domains) Every field is an integral domain, so all the
previous examples apply. In addition:
(a)
ℤ is an integral domain, but it is not a field.
(b)
ℝ[x] is not a field, since there is no polynomial P(x) with xP(x) = 1. However, ℝ[x]
is an integral domain, because if P(x)Q(x) = 0 then one of P or Q is zero.
(c)
ℤ[x] is also an example of an integral domain. In fact, R[x] is an integral domain for
any integral domain R (why?).
(d)
ℤ∕nℤ is a field (hence integral domain) exactly when n is prime. When n is not
prime, it is a ring but not an integral domain.
The trivial ring 0 is not considered an integral domain.
At this point, we go ahead and say:
Definition 5.2.5.An integral domain where all ideals are principal is called a principalideal domain (PID).
The ring ℤ∕6ℤ is an example of a ring which is a principal ideal ring, but not an integral
domain. As we alluded to earlier, we will never really use “principal ideal ring” in any real
way: we typically will want to strengthen it to PID.
5.3Prime ideals
Prototypical example for this section: (5) is a prime ideal of ℤ.
We know that every integer can be factored (up to sign) as a unique product of primes; for
example 15 = 3 ⋅ 5 and −10 = −2 ⋅ 5. You might remember the proof involves the so-called
Bézout’s lemma, which essentially says that (a,b) = (gcd(a,b)); in other words we’ve
carefully used the fact that ℤ is a PID.
It turns out that for general rings, the situation is not as nice as factoring elements
because most rings are not PID’s. The classic example of something going wrong
is
in ℤ[]. Nonetheless, we can sidestep the issue and talk about factoring ideals: somehow
the example 10 = 2 ⋅ 5 should be (10) = (2) ⋅ (5), which says “every multiple of 10 is the
product of a multiple of 2 and a multiple of 5”. I’d have to tell you then how to multiply two
ideals, which I do in the chapter on unique factorization.
Let’s at least figure out what primes are. In ℤ, we have that p≠1 is prime if whenever p∣xy,
either p∣x or p∣y. We port over this definition to our world of ideals.
Definition 5.3.1. A proper ideal I ⊊ R is a prime ideal if whenever xy ∈ I, either
x ∈ I or y ∈ I.
The condition that I is proper is analogous to the fact that we don’t consider 1 to be a prime
number.
Example 5.3.2 (Examples and non-examples of prime ideals)
(a)
The ideal (7) of ℤ is prime.
(b)
The ideal (8) of ℤ is not prime, since 2 ⋅ 4 = 8.
(c)
The ideal (x) of ℤ[x] is prime.
(d)
The ideal (x2) of ℤ[x] is not prime, since x ⋅ x = x2.
(e)
The ideal (3,x) of ℤ[x] is prime. This is actually easiest to see using ?? below.
(f)
The ideal (5) = 5ℤ +5iℤ of ℤ[i] is not prime, since the elements 3+i and 3−i have
product 10 ∈ (5), yet neither is itself in (5).
Remark 5.3.3 — Ideals have the nice property that they get rid of “sign issues”.
For example, in ℤ, do we consider −3 to be a prime? When phrased with ideals, this
annoyance goes away: (−3) = (3). More generally, for a ring R, talking about ideals lets
us ignore multiplication by a unit. (Note that −1 is a unit in ℤ.)
Exercise 5.3.4.What do you call a ring R for which the zero ideal (0) is prime?
We also have:
Theorem 5.3.5 (Prime ideal⇔quotient is integral domain) An ideal I is prime if and
only if R∕I is an integral domain.
Exercise 5.3.6 (Mandatory).Convince yourself the theorem is true; it is just definition chasing.(A possible start is to consider R = ℤ and I = (15).)
I now must regrettably inform you that unique factorization is still not true even with the
notion of a “prime” ideal (though again I haven’t told you how to multiply two ideals yet).
But it will become true with some additional assumptions that will arise in algebraic number
theory (relevant buzzword: Dedekind domain).
5.4Maximal ideals
Prototypical example for this section:The ideal (x,5) is maximal in ℤ[x], by quotient-ing.
Here’s another flavor of an ideal.
Definition 5.4.1.A proper ideal I of a ring R is maximal if it is not contained in any
other proper ideal.
Example 5.4.2 (Examples of maximal ideals)
(a)
The ideal I = (7) of ℤ is maximal, because if an ideal J contains 7 and an element
n not in I it must contain gcd(7,n) = 1, and hence J = ℤ.
(b)
The ideal (x) is not maximal in ℤ[x], because it’s contained in (x,5) (among others).
(c)
On the other hand, (x,5) is indeed maximal in ℤ[x]. This is actually easiest to verify
using ?? below.
(d)
Also, (x) is maximal in ℂ[x], again appealing to ?? below.
Exercise 5.4.3.What do you call a ring R for which the zero ideal (0) is maximal?
There’s an analogous theorem to the one for prime ideals.
Theorem 5.4.4 (I maximal⇔R∕I field) An ideal I is maximal if and only if R∕I is a
field.
Proof.A ring is a field if and only if (0) is the only maximal ideal. So this follows by
?? . □
Corollary 5.4.5 (Maximal ideals are prime) If I is a maximal ideal of a ring R, then I is
prime.
Proof.If I is maximal, then R∕I is a field, hence an integral domain, so I is prime. □
In practice, because modding out by generated ideals is pretty convenient, this is a very
efficient way to check whether an ideal is maximal.
Example 5.4.6 (Modding out in ℤ[x])
(a)
This instantly implies that (x,5) is a maximal ideal in ℤ[x], because if we mod out
by x and 5 in ℤ[x], we just get 𝔽5, which is a field.
(b)
On the other hand, modding out by just x gives ℤ, which is an integral domain but
not a field; that’s why (x) is prime but not maximal.
As we saw, any maximal ideal is prime. But now note that ℤ has the special property that
all of its nonzero prime ideals are also maximal. It’s with this condition and a few other minor
conditions that you get a so-called Dedekind domain where prime factorization of ideals does
work. More on that later.
5.5Field of fractions
Prototypical example for this section: Frac(ℤ) = ℚ.
As long as we are here, we take the time to introduce a useful construction that turns any
integral domain into a field.
Definition 5.5.1.Given an integral domain R, we define its field of fractions or
fraction field Frac(R) as follows: it consists of elements a∕b, where a,b ∈ R and
b≠0. We set a∕b ∼ c∕d if and only if bc = ad. Addition and multiplication is defined
by
+
=
⋅
= .
In fact everything you know about ℚ basically carries over by analogy. You can prove if you
want that this indeed a field, but considering how comfortable we are that ℚ is well-defined, I
wouldn’t worry about it…
Definition 5.5.2.Let k be a field. We define k(x) = Frac(k[x]) (read “k of x”), and
call it the field of rational functions.
Example 5.5.3 (Examples of fraction fields)
(a)
By definition, Frac(ℤ) = ℚ.
(b)
The field ℝ(x) consists of rational functions in x:
For example, might be a typical element.
Example 5.5.4 (Gaussian rationals) Just like we defined ℤ[i] by abusing notation, we
can also write ℚ(i) = Frac(ℤ[i]). Officially, it should consist of
for polynomials f and g with rational coefficients. But since i2 = −1 this just leads to
And since = we end up with
5.6Unique factorization domains (UFD’s)
Prototypical example for this section:ℤ and polynomial rings in general.
Here is one stray definition that will be important for those with a number-theoretic
inclination. Over the positive integers, we have a fundamental theorem of arithmetic, stating
that every integer is uniquely the product of prime numbers.
We can even make an analogous statement in ℤ or ℤ[i], if we allow representations like
6 = (−2)(−3) and so on. The trick is that we only consider everything up to units; so
6 = (−2)(−3) = 2 ⋅ 3 are considered the same.
The general definition goes as follows.
Definition 5.6.1.A nonzero non-unit of an integral domain R is irreducible if it
cannot be written as the product of two non-units.
An integral domain R is a unique factorization domain if every nonzero non-unit
of R can be written as the product of irreducible elements, which is unique up to
multiplication by units.
Question 5.6.2.Verify that ℤ is a UFD.
Example 5.6.3 (Examples of UFD’s)
(a)
Fields are a “degenerate” example of UFD’s: every nonzero element is a unit, so
there is nothing to check.
(b)
ℤ is a UFD. The irreducible elements are p and −p, for example 5 or −17.
(c)
ℚ[x] is a UFD: polynomials with rational coefficients can be uniquely factored, up
to scaling by constants (as the units of ℚ[x] are just the rational numbers).
(d)
ℤ[x] is a UFD.
(e)
The Gaussian integers ℤ[i] turns out to be a UFD too (and this will be proved in
the chapters on algebraic number theory).
(f)
ℤ[] is the classic non-example of a UFD: one may write
but each of 2, 3, 1 ± is irreducible. (It turns out the right way to fix this is by
considering prime ideals instead, and this is one big motivation for ?? .)
(g)
Theorem we won’t prove: every PID is a UFD.
(h)
Theorem we won’t prove: if R is a UFD, so is R[x] (and hence by induction so is
R[x,y], R[x,y,z], …).
5.7A few harder problems to think about
Not olympiad problems, but again the spirit is very close to what you might see in an
olympiad.
Problem 5A.Consider the ring
Is it a field?
Problem 5B (Homomorphisms from fields are injective). Let K be
a field and R a ring. Prove that any homomorphism ψ: K → R is
injective.2
Problem 5C⋆ (Pre-image of prime ideals). Suppose ϕ: R → S is a ring homomorphism, and
I ⊆ S is a prime ideal. Prove that ϕpre(I) is prime as well.
Problem 5D⋆.Let R be an integral domain with finitely many elements.
Prove that R is a field.
Problem 5E⋆ (Krull’s theorem). Let R be a ring and J a proper ideal.
(a)
Prove that if R is Noetherian, then J is contained in a maximal ideal I.
(b)
Use Zorn’s lemma (?? ) to prove the result even if R isn’t Noetherian.
Problem 5F (Speck[x]).Describe the prime ideals of ℂ[x] and ℝ[x].
Problem 5G.Prove that any nonzero prime ideal of ℤ[] is also a maximal ideal.
Part III Basic Topology
6Properties of metric spaces
At the end of the last chapter on metric spaces, we introduced two adjectives “open” and
“closed”. These are important because they’ll grow up to be the definition for a general
topological space, once we graduate from metric spaces.
To move forward, we provide a couple niceness adjectives that applies to entire metricspaces, rather than just a set relative to a parent space. They are “(totally) bounded” and
“complete”. These adjectives are specific to metric spaces, but will grow up to become the
notion of compactness, which is, in the words of [?], “the single most important concept
in real analysis”. At the end of the chapter, we will know enough to realize that
something is amiss with our definition of homeomorphism, and this will serve as
the starting point for the next chapter, when we define fully general topological
spaces.
6.1Boundedness
Prototypical example for this section: [0,1] is bounded but ℝ is not.
Here is one notion of how to prevent a metric space from being a bit too large.
Definition 6.1.1.A metric space M is bounded if there is a constant D such that
d(p,q) ≤ D for all p,q ∈ M.
You can change the order of the quantifiers:
Proposition 6.1.2 (Boundedness with radii instead of diameters) A metric space M is
bounded if and only if for every point p ∈ M, there is a radius R (possibly depending
on p) such that d(p,q) ≤ R for all q ∈ M.
Exercise 6.1.3.Use the triangle inequality to show these are equivalent. (The names “radius”and “diameter” are a big hint!)
Example 6.1.4 (Examples of bounded spaces)
(a)
Finite intervals like [0,1] and (a,b) are bounded.
(b)
The unit square [0,1]2 is bounded.
(c)
ℝn is not bounded for any n ≥ 1.
(d)
A discrete space on an infinite set is bounded.
(e)
ℕ is not bounded, despite being homeomorphic to the discrete space!
The fact that a discrete space on an infinite set is “bounded” might be upsetting to you, so
here is a somewhat stronger condition you can use:
Definition 6.1.5.A metric space is totally bounded if for any 𝜀 > 0, we can cover
M with finitely many 𝜀-neighborhoods.
For example, if 𝜀 = 1∕2, you can cover [0,1]2 by 𝜀-neighborhoods.
Exercise 6.1.6.Show that “totally bounded” implies “bounded”.
Example 6.1.7 (Examples of totally bounded spaces)
(a)
A subset of ℝn is bounded if and only if it is totally bounded.
This is for Euclidean geometry reasons: for example in ℝ2 if I can cover a set by a
single disk of radius 2, then I can certainly cover it by finitely many disks of radius
1∕2. (We won’t prove this rigorously.)
(b)
So for example [0,1] or [0,2] × [0,3] is totally bounded.
(c)
In contrast, a discrete space on an infinite set is not totally bounded.
6.2Completeness
Prototypical example for this section:ℝ is complete, but ℚ and (0,1) are not.
So far we can only talk about sequences converging if they have a limit. But consider the
sequence
It converges to in ℝ, of course. But it fails to converge in ℚ; there is no rational
number this sequence converges to. And so somehow, if we didn’t know about the
existence of ℝ, we would have no idea that the sequence (xn) is “approaching”
something.
That seems to be a shame. Let’s set up a new definition to describe these sequences whose
terms get close to each other, even if they don’t approach any particular point in the space.
Thus, we only want to mention the given points in the definition.
Definition 6.2.1.Let x1,x2,… be a sequence which lives in a metric space M =
(M,dM). We say the sequence is Cauchy if for any 𝜀 > 0, we have
for all sufficiently large m and n.
Question 6.2.2.Show that a sequence which converges is automatically Cauchy. (Draw apicture.)
Now we can define:
Definition 6.2.3.A metric space M is complete if every Cauchy sequence converges.
Example 6.2.4 (Examples of complete spaces)
(a)
ℝ is complete. (Depending on your definition of ℝ, this either follows by definition,
or requires some work. We won’t go through this here.)
(b)
The discrete space is complete, as the only Cauchy sequences are eventually constant.
(c)
The closed interval [0,1] is complete.
(d)
ℝn is complete as well. (You’re welcome to prove this by induction on n.)
Example 6.2.5 (Non-examples of complete spaces)
(a)
The rationals ℚ are not complete.
(b)
The open interval (0,1) is not complete, as the sequence 0.9, 0.99, 0.999, 0.9999, …is
Cauchy but does not converge.
So, metric spaces need not be complete, like ℚ. But we certainly would like them to be
complete, and in light of the following theorem this is not unreasonable.
Theorem 6.2.6 (Completion) Every metric space can be “completed”, i.e. made into a
complete space by adding in some points.
We won’t need this construction at all, so it’s left as ?? .
Example 6.2.7 (ℚ completes to ℝ) The completion of ℚ is ℝ.
(In fact, by using a modified definition of completion not depending on the real numbers,
other authors often use this as the definition of ℝ.)
6.3Let the buyer beware
There is something suspicious about both these notions: neither are preserved under
homeomorphism!
Example 6.3.1 (Something fishy is going on here) Let M = (0,1) and N = ℝ. As we saw
much earlier M and N are homeomorphic. However:
(0,1) is totally bounded, but not complete.
ℝ is complete, but not bounded.
This is the first hint of something going awry with the metric. As we progress further into
our study of topology, we will see that in fact open and closed sets (which we motivated by
using the metric) are the notion that will really shine later on. I insist on introducing the
metric first so that the standard pictures of open and closed sets make sense, but eventually it
becomes time to remove the training wheels.
6.4Subspaces, and (inb4) a confusing linguistic point
Prototypical example for this section:A circle is obtained as a subspace of ℝ2.
As we’ve already been doing implicitly in examples, we’ll now say:
Definition 6.4.1.Every subset S ⊆ M is a metric space in its own right, by re-using
the distance function on M. We say that S is a subspace of M.
For example, we saw that the circle S1 is just a subspace of ℝ2.
It thus becomes important to distinguish between
(i)
“absolute” adjectives like “complete” or “bounded”, which can be applied to
both spaces, and hence even to subsets of spaces (by taking a subspace), and
(ii)
“relative” adjectives like “open (in M)” and “closed (in M)”, which make sense
only relative to a space, even though people are often sloppy and omit them.
So “[0,1] is complete” makes sense, as does “[0,1] is a complete subset of ℝ”, which we take to
mean “[0,1] is a complete as a subspace of ℝ”. This is since “complete” is an absolute
adjective.
But here are some examples of ways in which relative adjectives require a little more care:
Consider the sequence 1, 1.4, 1.41, 1.414, …. Viewed as a sequence in ℝ, it converges
to . But if viewed as a sequence in ℚ, this sequence does not converge! Similarly,
the sequence 0.9, 0.99, 0.999, 0.9999 does not converge in the space (0,1), although
it does converge in [0,1].
The fact that these sequences fail to converge even though they “ought to” is weird
and bad, and was why we defined complete spaces to begin with.
In general, it makes no sense to ask a question like “is [0,1] open?”. The questions
“is [0,1] open in ℝ?” and “is [0,1] open in [0,1]?” do make sense, however. The
answer to the first question is “no” but the answer to the second question is “yes”;
indeed, every space is open in itself. Similarly, [0,) is an open set in the space
M = [0,1] because it is the ball in M of radius centered at 0.
Dually, it doesn’t make sense to ask “is [0,1] closed”? It is closed in ℝ and in itself
(but every space is closed in itself, anyways).
To make sure you understand the above, here are two exercises to help you practice relative
adjectives.
Exercise 6.4.2.Let M be a complete metric space and let S ⊆ M. Prove that S is complete ifand only if it is closed in M. In particular, [0,1] is complete.
Exercise 6.4.3.Let M = [0,1] ∪ (2,3). Show that [0,1] and (2,3) are both open and closed inM.
This illustrates a third point: a nontrivial set can be both open and
closed1
As we’ll see in ?? , this implies the space is disconnected; i.e. the only examples look quite
like the one we’ve given above.
6.5A few harder problems to think about
Problem 6A† (Banach fixed point theorem).Let M = (M,d) be a complete metric
space. Suppose T : M → M is a continuous map such that for any p≠q ∈ M,
(We call T a contraction.) Show that T has a unique fixed point.
Problem 6B (Henning Makholm, on math.SE).We let M and N denote the metric spaces
obtained by equipping ℝ with the following two metrics:
dM(x,y)
= min
dN(x,y)
= .
(a)
Fill in the following 2 × 3 table with “yes” or “no” for each cell.
Complete?
Bounded?
Totally bounded?
M
N
(b)
Are M and N homeomorphic?
Problem 6C† (Completion of a metric space). Let M be a metric space.
Construct a complete metric space M such that M is a subspace of M, and every open set of
M contains a point of M (meaning M is dense in M).
Problem 6D.Show that a metric space is totally bounded if and only if any sequence has a
Cauchy subsequence.
Problem 6E.Prove that ℚ is not homeomorphic to any complete metric
space.
7Topological spaces
In ?? we introduced the notion of space by describing metrics on them. This gives you a lot
examples, and nice intuition, and tells you how you should draw pictures of open and closed
sets.
However, moving forward, it will be useful to begin thinking about topological
spaces in terms of just their open sets. (One motivation is that our fishy ?? shows
that in some ways the notion of homeomorphism really wants to be phrased
in terms of open sets, not in terms of the metric.) As we are going to see, the
open sets manage to actually retain nearly all the information we need, but are
simpler.1
This will be done in just a few sections, and after that we will start describing more adjectives
that we can apply to topological (and hence metric) spaces.
The most important topological notion is missing from this chapter: that of a
compact space. It is so important that I have dedicated a separate chapter just for
it.
Quick note for those who care: the adjectives “Hausdorff”, “connected”, and later
“compact” are all absolute adjectives.
7.1Forgetting the metric
Recall ?? :
A function f : M → N of metric spaces is continuous if and only if the
pre-image of every open set in N is open in M.
Despite us having defined this in the context of metric spaces, this nicely doesn’t refer to the
metric at all, only the open sets. As alluded to at the start of this chapter, this is a great
motivation for how we can forget about the fact that we had a metric to begin with, and
rather start with the open sets instead.
Definition 7.1.1.A topological space is a pair (X,𝒯 ), where X is a set of points, and 𝒯 is
the topology, which consists of several subsets of X, called the open sets of X. The
topology must obey the following axioms.
∅ and X are both in 𝒯 .
Finite intersections of open sets are also in 𝒯 .
Arbitrary unions (possibly infinite) of open sets are also in 𝒯 .
So this time, the open sets are given. Rather than defining a metric and getting open sets from
the metric, we instead start from just the open sets.
Abuse of Notation 7.1.2.We abbreviate (X,𝒯 ) by just X, leaving the topology 𝒯
implicit. (Do you see a pattern here?)
Example 7.1.3 (Examples of topologies)
(a)
Given a metric space M, we can let 𝒯 be the open sets in the metric sense. The
point is that the axioms are satisfied.
(b)
In particular, discrete space is a topological space in which every set is open.
(Why?)
(c)
Given X, we can let 𝒯 = , the opposite extreme of the discrete space.
Now we can port over our metric definitions.
Definition 7.1.4.An open neighborhood2
of a point x ∈ X is an open set U which contains x (see figure).
Abuse of Notation 7.1.5.Just to be perfectly clear: by an “open neighborhood” I
mean any open set containing x. But by an “r-neighborhood” I always mean the points
with distance less than r from x, and so I can only use this term if my space is a metric
space.
7.2Re-definitions
Now that we’ve defined a topological space, for nearly all of our metric notions we can write
down as the definition the one that required only open sets (which will of course agree with
our old definitions when we have a metric space).
7.2.iContinuity
Here was our motivating example, continuity:
Definition 7.2.1.We say function f : X → Y of topological spaces is continuous at a
point p ∈ X if the pre-image of any open neighborhood of f(p) is an open neighborhood
of p. The function is continuous if it is continuous at every point.
Thus homeomorphisms carries over: a bijection which is continuous in both directions.
Definition 7.2.2.A homeomorphism of topological spaces (X,τX) and (Y,τY) is a
bijection f : X → Y which induces a bijection from τX to τY: i.e. the bijection preserves
open sets.
Question 7.2.3.Show that this is equivalent to f and its inverse both being continuous.
Therefore, any property defined only in terms of open sets is preserved by homeomorphism.
Such a property is called a topological property. However, the later adjectives we define
(“connected”, “Hausdorff”, “compact”) will all be defined only in terms of the open sets, so
they will be.
7.2.iiClosed sets
We saw last time there were two equivalent definitions for closed sets, but one of them relies
only on open sets, and we use it:
Definition 7.2.4. In a general topological space X, we say that S ⊆ X is closed in
X if the complement X ∖ S is open in X.
If S ⊆ X is any set, the closure of S, denoted S, is defined as the smallest closed set
containing S.
Thus for general topological spaces, open and closed sets carry the same information, and it is
entirely a matter of taste whether we define everything in terms of open sets or closed
sets. In particular, you can translate axioms and properties of open sets to closed
ones:
Question 7.2.5.Show that the (possibly infinite) intersection of closed sets is closed while theunion of finitely many closed sets is closed. (Look at complements.)
Exercise 7.2.6.Show that a function is continuous if and only if the pre-image of every closedset is closed.
Mathematicians seem to have agreed that they like open sets better.
7.2.iiiProperties that don’t carry over
Not everything works:
Remark 7.2.7 (Complete and (totally) bounded are metric properties) — The two
metric properties we have seen, “complete” and “(totally) bounded”, are not topological
properties. They rely on a metric, so as written we cannot apply them to topological
spaces. One might hope that maybe, there is some alternate definition (like we saw for
“continuous function”) that is just open-set based. But ?? showing (0,1)ℝ tells us
that it is hopeless.
Remark 7.2.8 (Sequences don’t work well) — You could also try to port over the notion
of sequences and convergent sequences. However, this turns out to break a lot of desirable
properties. Therefore I won’t bother to do so, and thus if we are discussing sequences
you should assume that we are working with a metric space.
7.3Hausdorff spaces
Prototypical example for this section:Every space that’s not the Zariski topology (defined muchlater).
As you might have guessed, there exist topological spaces which cannot be realized as metric
spaces (in other words, are not metrizable). One example is just to take X = {a,b,c} and the
topology τX = . This topology is fairly “stupid”: it can’t tell apart any of the
points a, b, c! But any metric space can tell its points apart (because d(x,y) > 0 when
x≠y).
We’ll see less trivial examples later, but for now we want to add a little more sanity
condition onto our spaces. There is a whole hierarchy of such axioms, labelled Tn for integers
n (with n = 0 being the weakest and n = 6 the strongest); these axioms are called separationaxioms.
By far the most common hypothesis is the T2 axiom, which bears a special name.
Definition 7.3.1.A topological space X is Hausdorff if for any two distinct points p
and q in X, there exists an open neighborhood U of p and an open neighborhood V of
q such that
In other words, around any two distinct points we should be able to draw disjoint
open neighborhoods. Here’s a picture to go with above, but not much going on.
Question 7.3.2.Show that all metric spaces are Hausdorff.
I just want to define this here so that I can use this word later. In any case, basically any
space we will encounter other than the Zariski topology is Hausdorff.
7.4Subspaces
Prototypical example for this section:S1is a subspace of ℝ2.
One can also take subspaces of general topological spaces.
Definition 7.4.1.Given a topological space X, and a subset S ⊆ X, we can make S
into a topological space by declaring that the open subsets of S are U ∩ S for open
U ⊆ X. This is called the subspace topology.
So for example, if we view S1 as a subspace of ℝ2, then any open arc is an open set, because
you can view it as the intersection of an open disk with S1.
Needless to say, for metric spaces it doesn’t matter which of these definitions I choose.
(Proving this turns out to be surprisingly annoying, so I won’t do so.)
7.5Connected spaces
Prototypical example for this section: [0,1] ∪ [2,3] is disconnected.
Even in metric spaces, it is possible for a set to be both open and closed.
Definition 7.5.1.A subset S of a topological space X is clopen if it is both closed
and open in X. (Equivalently, both S and its complement are open.)
For example ∅ and the entire space are examples of clopen sets. In fact, the presence
of a nontrivial clopen set other than these two leads to a so-called disconnected
space.
Question 7.5.2.Show that a space X has a nontrivial clopen set (one other than ∅ and X) ifand only if X can be written as a disjoint union of two nonempty open sets.
We say X is disconnected if there are nontrivial clopen sets, and connected
otherwise. To see why this should be a reasonable definition, it might help to solve
?? .
Example 7.5.3 (Disconnected and connected spaces)
(a)
The metric space
is disconnected (it consists of two disks).
(b)
The space [0,1] ∪ [2,3] is disconnected: it consists of two segments, each of which is
a clopen set.
(c)
A discrete space on more than one point is disconnected, since every set is clopen
in the discrete space.
(d)
Convince yourself that the set
is a clopen subset of ℚ. Hence ℚ is disconnected too – it has gaps.
(e)
[0,1] is connected.
7.6Path-connected spaces
Prototypical example for this section:Walking around in ℂ.
A stronger and perhaps more intuitive notion of a connected space is a path-connected
space. The short description: “walk around in the space”.
Definition 7.6.1.A path in the space X is a continuous function
Its endpoints are the two points γ(0) and γ(1).
You can think of [0,1] as measuring “time”, and so we’ll often write γ(t) for t ∈ [0,1] (with t
standing for “time”). Here’s a picture of a path.
Question 7.6.2.Why does this agree with your intuitive notion of what a “path” is?
Definition 7.6.3.A space X is path-connected if any two points in it are connected
by some path.
Exercise 7.6.4 (Path-connected implies connected).Let X = U ⊔ V be a disconnected space.Show that there is no path from a point of U to point V . (If γ:[0,1] → X, then we get[0,1] = γpre(U) ⊔ γpre(V ), but [0,1] is connected.)
Example 7.6.5 (Examples of path-connected spaces)
ℝ2 is path-connected, since we can “connect” any two points with a straight line.
The unit circle S1 is path-connected, since we can just draw the major or minor
arc to connect two points.
7.7Homotopy and simply connected spaces
Prototypical example for this section:ℂ and ℂ ∖{0}.
Now let’s motivate the idea of homotopy. Consider the example of the complex plane ℂ
(which you can think of just as ℝ2) with two points p and q. There’s a whole bunch of paths
from p to q but somehow they’re not very different from one another. If I told you “walk from
p to q” you wouldn’t have too many questions.
So we’re living happily in ℂ until a meteor strikes the origin, blowing it out of existence.
Then suddenly to get from p to q, people might tell you two different things: “go left around
the meteor” or “go right around the meteor”.
So what’s happening? In the first picture, the red, green, and blue paths somehow all looked
the same: if you imagine them as pieces of elastic string pinned down at p and q, you can
stretch each one to any other one.
But in the second picture, you can’t move the red string to match
with the blue string: there’s a meteor in the way. The paths are actually
different.3
The formal notion we’ll use to capture this is homotopy equivalence. We want to write a
definition such that in the first picture, the three paths are all homotopic, but the two paths in
the second picture are somehow not homotopic. And the idea is just continuous
deformation.
Definition 7.7.1.Let α and β be paths in X whose endpoints coincide. A (path)
homotopy from α to β is a continuous function F : [0,1]2→ X, which we’ll write Fs(t)
for s,t ∈ [0,1], such that
and moreover
If a path homotopy exists, we say α and β are path homotopic and write α ≃ β.
Abuse of Notation 7.7.2.While I strictly should say “path homotopy” to describe
this relation between two paths, I will shorten this to just “homotopy” instead. Similarly
I will shorten “path homotopic” to “homotopic”.
What this definition is doing is taking α and “continuously deforming” it to β, while
keeping the endpoints fixed. Note that for each particular s, Fs is itself a function. So s
represents time as we deform α to β: it goes from 0 to 1, starting at α and ending at
β.
Question 7.7.3.Convince yourself the above definition is right. What goes wrong when themeteor strikes?
So now I can tell you what makes ℂ special:
Definition 7.7.4.A space X is simply connected if it’s path-connected and for any
points p and q, all paths from p to q are homotopic.
That’s why you don’t ask questions when walking from p to q in ℂ: there’s really only one way
to walk. Hence the term “simply” connected.
Question 7.7.5.Convince yourself that ℝnis simply connected for all n.
7.8Bases of spaces
Prototypical example for this section:ℝ has a basis of open intervals, and ℝ2has a basis ofopen disks.
You might have noticed that the open sets of ℝ are a little annoying to describe: the
prototypical example of an open set is (0,1), but there are other open sets like
Question 7.8.1.Check this is an open set.
But okay, this isn’t that different. All I’ve done is taken a bunch of my prototypes and threw
a bunch of ∪ signs at it. And that’s the idea behind a basis.
Definition 7.8.2.A basis for a topological space X is a subset ℬ of the open sets such
that every open set in X is a union of some (possibly infinite) number of elements in ℬ.
And all we’re doing is saying
Example 7.8.3 (Basis of ℝ) The open intervals form a basis of ℝ.
In fact, more generally we have:
Theorem 7.8.4 (Basis of metric spaces) The r-neighborhoods form a basis of any metric
space M.
Proof.Kind of silly – given an open set U draw an rp-neighborhood Up contained entirely
inside U. Then ⋃pUp is contained in U and covers every point inside it. □
Hence, an open set in ℝ2 is nothing more than a union of a bunch of open disks, and so on.
The point is that in a metric space, the only open sets you really ever have to worry too much
about are the r-neighborhoods.
7.9A few harder problems to think about
Problem 7A†.Let X be a topological space. Show that there exists a nonconstant
continuous function X →{0,1} if and only if X is disconnected (here {0,1} is given the
discrete topology).
Problem 7B⋆.Let X and Y be topological spaces and let f : X → Y be a continuous
function.
(a)
Show that if X is connected then so is fimg(X).
(b)
Show that if X is path-connected then so is fimg(X).
Problem 7C (Hausdorff implies T1 axiom).Let X be a Hausdorff topological space.
Prove that for any point p ∈ X the set {p} is closed.
Problem 7D ([?], Exercise 2.56).Let M be a metric space with more than one
point but at most countably infinitely many points. Show that M is disconnected.
Problem 7E (Furstenberg).We declare a subset of ℤ to be open if it’s the union (possibly
empty or infinite) of arithmetic sequences , where a and d are positive
integers.
(a)
Verify this forms a topology on ℤ, called the evenly spaced integer topology.
(b)
Prove there are infinitely many primes by considering ⋃ppℤ for primes p.
Problem 7F.Prove that the evenly spaced integer topology on ℤ is
metrizable. In other words, show that one can impose a metric d : ℤ2→ ℝ which makes ℤ into
a metric space whose open sets are those described above.
Problem 7G.We know that any open set U ⊆ ℝ is a
union of open intervals (allowing ±∞ as endpoints). One can show that
it’s actually possible to write U as the union of pairwise disjoint open
intervals.4
Prove that there exists such a disjoint union with at most countably many intervals in it.
8Compactness
One of the most important notions of topological spaces is that of compactness. It generalizes
the notion of “closed and bounded” in Euclidean space to any topological space (e.g. see
?? ).
For metric spaces, there are two equivalent ways of formulating compactness:
A “natural” definition using sequences, called sequential compactness.
A less natural definition using open covers.
As I alluded to earlier, sequences in metric spaces are super nice, but sequences in general
topological spaces suck (to the point where I didn’t bother to define convergence
of general sequences). So it’s the second definition that will be used for general
spaces.
8.1Definition of sequential compactness
Prototypical example for this section: [0,1] is compact, but (0,1) is not.
To emphasize, compactness is one of the best possible properties that a metric space can
have.
Definition 8.1.1.A subsequence of an infinite sequence x1,x2,… is exactly what it
sounds like: a sequence xi1,xi2,… where i1< i2< are positive integers. Note that
the sequence is required to be infinite.
Another way to think about this is “selecting infinitely many terms” or “deleting
some terms” of the sequence, depending on whether your glass is half empty or half
full.
Definition 8.1.2.A metric space M is sequentially compact if every sequence has
a subsequence which converges.
This time, let me give some non-examples before the examples.
Example 8.1.3 (Non-examples of compact metric spaces)
(a)
The space ℝ is not compact: consider the sequence 1,2,3,4,…. Any subsequence
explodes, hence ℝ cannot possibly be compact.
(b)
More generally, if a space is not bounded it cannot be compact. (You can prove this
if you want.)
(c)
The open interval (0,1) is bounded but not compact: consider the sequence ,,,….
No subsequence can converge to a point in (0,1) because the sequence “converges
to 0”.
(d)
More generally, any space which is not complete cannot be compact.
Now for the examples!
Question 8.1.4.Show that a finite set is compact. (Pigeonhole Principle.)
Example 8.1.5 (Examples of compact spaces) Here are some more examples of compact
spaces. I’ll prove they’re compact in just a moment; for now just convince yourself they
are.
(a)
[0,1] is compact. Convince yourself of this! Imagine having a large number of dots
in the unit interval…
(b)
The surface of a sphere, S2 = is compact.
(c)
The unit ball B2 = is compact.
(d)
The Hawaiian earring living in ℝ2 is compact: it consists of mutually tangent
circles of radius for each n, as in ?? .
Figure 8.1:Hawaiian Earring.
To aid in generating more examples, we remark:
Proposition 8.1.6 (Closed subsets of compacts) Closed subsets of sequentially compact
sets are compact.
Question 8.1.7.Prove this. (It should follow easily from definitions.)
We need to do a bit more work for these examples, which we do in the next section.
8.2Criteria for compactness
Theorem 8.2.1 (Tychonoff’s theorem) If X and Y are compact spaces, then so is X ×Y .
Proof.?? . □
We also have:
Theorem 8.2.2 (The interval is compact) [0,1] is compact.
Proof.Killed by ?? ; however, here is a sketch of a direct proof. Split [0,1] into
[0,] ∪ [,1]. By Pigeonhole, infinitely many terms of the sequence lie in the left half
(say); let x1 be the first one and then keep only the terms in the left half after x1. Now
split [0,] into [0,] ∪ [,]. Again, by Pigeonhole, infinitely many terms fall in some
half; pick one of them, call it x2. Rinse and repeat. In this way we generate a sequence
x1, x2, …which is Cauchy, implying that it converges since [0,1] is complete. □
Now we can prove the main theorem about Euclidean space: in ℝn, compactness is
equivalent to being “closed and bounded”.
Theorem 8.2.3 (Bolzano-Weierstraß) A subset of ℝn is compact if and only if it is closed
and bounded.
Question 8.2.4.Why does this imply the spaces in our examples are compact?
Proof.Well, look at a closed and bounded S ⊆ ℝn. Since it’s bounded, it lives inside
some box [a1,b1] × [a2,b2] ×× [an,bn]. By Tychonoff’s theorem, since each [ai,bi] is
compact the entire box is. Since S is a closed subset of this compact box, we’re done. □
One really has to work in ℝn for this to be true! In other spaces, this criterion can easily
fail.
Example 8.2.5 (Closed and bounded but not compact) Let S = {s1,s2,…} be any
infinite set equipped with the discrete metric. Then S is closed (since all convergent
sequences are constant sequences) and S is bounded (all points are a distance 1 from
each other) but it’s certainly not compact since the sequence s1,s2,… doesn’t converge.
The Bolzano-Weierstrass theorem, which is ?? , tells you exactly which sets are compact in
metric spaces in a geometric way.
8.3Compactness using open covers
Prototypical example for this section: [0,1] is compact.
There’s a second related notion of compactness which I’ll now define. The following
definitions might appear very unmotivated, but bear with me.
Definition 8.3.1.An open cover of a topological space X is a collection of open sets
{Uα} (possibly infinite or uncountable) which cover it: every point in X lies in at least
one of the Uα, so that
Such a cover is called an open cover.
A subcover is exactly what it sounds like: it takes only some of the Uα, while ensuring
that X remains covered.
Some art:
Definition 8.3.2.A topological space X is quasicompact if every open cover has a
finite subcover. It is compact if it is also Hausdorff.
Remark 8.3.3 — The “Hausdorff” hypothesis that I snuck in is a sanity condition which
is not worth worrying about unless you’re working on the algebraic geometry chapters,
since all the spaces you will deal with are Hausdorff. (In fact, some authors don’t even
bother to include it.) For example all metric spaces are Hausdorff and thus this condition
can be safely ignored if you are working with metric spaces.
What does this mean? Here’s an example:
Example 8.3.4 (Example of a finite subcover) Suppose we cover the unit square M =
[0,1]2 by putting an open disk of diameter 1 centered at every point (trimming any
overflow). This is clearly an open cover because, well, every point lies in many of the
open sets, and in particular is the center of one.
But this is way overkill – we only need about four of these circles to cover the whole square.
That’s what is meant by a “finite subcover”.
Why do we care? Because of this:
Theorem 8.3.5 (Sequentially compact⇔compact) A metric space M is sequentially
compact if and only if it is compact.
We defer the proof to the last section.
This gives us the motivation we wanted for our definition. Sequential compactness was a
condition that made sense. The open-cover definition looked strange, but it turned out to be
equivalent. But we now prefer it, because we have seen that whenever possible we want
to resort to open-set-only based definitions: so that e.g. they are preserved under
homeomorphism.
Example 8.3.6 (An example of non-compactness) The space X = [0,1) is not compact
in either sense. We can already see it is not sequentially compact, because it is not even
complete (look at xn = 1 −). To see it is not compact under the covering definition,
consider the sets
for m = 1,2,…. Then X = ⋃Ui; hence the Ui are indeed a cover. But no finite collection
of the Ui’s will cover X.
Question 8.3.7.Convince yourself that [0,1] is compact; this is a little less intuitive than itbeing sequentially compact.
Abuse of Notation 8.3.8.Thus, we’ll never call a metric space “sequentially
compact” again — we’ll just say “compact”. (Indeed, I kind of already did this in the
previous few sections.)
8.4Applications of compactness
Compactness lets us reduce infinite open covers to finite ones. Actually, it lets us do
this even if the open covers are blithely stupid. Very often one takes an open cover
consisting of an open neighborhood of x ∈ X for every single point x in the space;
this is a huge number of open sets, and yet compactness lets us reduce to a finite
set.
To give an example of a typical usage:
Proposition 8.4.1 (Compacttotally bounded) Let M be compact. Then M is
totally bounded.
Proof using covers. For every point p ∈ M, take an 𝜀-neighborhood of p, say Up. These
cover M for the horrendously stupid reason that each point p is at the very least covered
by its open neighborhood Up. Compactness then lets us take a finite subcover. □
Next, an important result about maps between compact spaces.
Theorem 8.4.2 (Images of compacts are compact) Let f : X → Y be a continuous
function, where X is compact. Then the image
is compact.
Proof using covers. Take any open cover {Vα} in Y of fimg(X). By continuity of f, it
pulls back to an open cover {Uα} of X. Thus some finite subcover of this covers X. The
corresponding V ’s cover fimg(X). □
Question 8.4.3.Give another proof using the sequential definitions of continuity andcompactness. (This is even easier.)
Some nice corollaries of this:
Corollary 8.4.4 (Extreme value theorem) Let X be compact and consider a continuous
function f : X → ℝ. Then f achieves a maximum value at some point, i.e. there is a
point p ∈ X such that f(p) ≥ f(q) for any other q ∈ X.
Corollary 8.4.5 (Intermediate value theorem) Consider a continuous function f : [0,1] →ℝ. Then the image of f is of the form [a,b] for some real numbers a ≤ b.
Sketch of Proof. The point is that the image of f is compact in ℝ, and hence closed
and bounded. You can convince yourself that the closed sets are just unions of closed
intervals. That implies the extreme value theorem.
When X = [0,1], the image is also connected, so there should only be one closed
interval in fimg([0,1]). Since the image is bounded, we then know it’s of the form [a,b].
(To give a full proof, you would use the so-called least upper bound property, but that’s
a little involved for a bedtime story; also, I think ℝ is boring.) □
Example 8.4.6 (1∕x) The compactness hypothesis is really important here. Otherwise,
consider the function
This function (which you plot as a hyperbola) is not bounded; essentially, you can see
graphically that the issue is we can’t extend it to a function on [0,1] because it explodes
near x = 0.
One last application: if M is a compact metric space, then continuous functions f : M → N
are continuous in an especially “nice” way:
Definition 8.4.7.A function f : M → N of metric spaces is called uniformlycontinuous if for any 𝜀 > 0, there exists a δ > 0 (depending only on 𝜀) such that
whenever dM(x,y) < δ we also have dN(fx,fy) < 𝜀.
The name means that for 𝜀 > 0, we need a δ that works for every point of M.
Example 8.4.8 (Uniform continuity)
(a)
The functions ℝ to ℝ of the form xax + b are all uniformly continuous, since one
can always take δ = 𝜀∕|a| (or δ = 1 if a = 0).
(b)
Actually, it is true that a differentiable function ℝ → ℝ with a bounded derivative is
uniformly continuous. (The converse is false for the reason that uniformly continuous
doesn’t imply differentiable at all.)
(c)
The function f : ℝ → ℝ by xx2 is not uniformly continuous, since for large x,
tiny δ changes to x lead to fairly large changes in x2. (If you like, you can try to
prove this formally now.)
Think f(2017.01) − f(2017) > 40; even when δ = 0.01, one can still cause large
changes in f.
(d)
However, when restricted to (0,1) or [0,1] the function xx2 becomes uniformly
continuous. (For 𝜀 > 0 one can now pick for example δ = min{1,𝜀}∕3.)
(e)
The function (0,1) → ℝ by x1∕x is not uniformly continuous (same reason as
before).
Now, as promised:
Proposition 8.4.9 (Continuous on compactuniformly continuous) If M is compact
and f : M → N is continuous, then f is uniformly continuous.
Proof using sequences. Fix 𝜀 > 0, and assume for contradiction that for every δ = 1∕k
there exists points xk and yk within δ of each other but with images 𝜀 > 0 apart. By
compactness, take a convergent subsequence xik→ p. Then yik→ p as well, since the
xk’s and yk’s are close to each other. So both sequences f(xik) and f(yik) should converge
to f(p) by sequential continuity, but this can’t be true since the two sequences are always
𝜀 apart. □
8.5(Optional) Equivalence of formulations of compactness
We will prove that:
Theorem 8.5.1 (Heine-Borel for general metric spaces) For a metric space M, the following are
equivalent:
(i)
Every sequence has a convergent subsequence,
(ii)
The space M is complete and totally bounded, and
(iii)
Every open cover has a finite subcover.
We leave the proof that (i) ⇔ (ii) as ?? ; the idea of the proof is much in the spirit of
?? .
Proof that (i) and (ii)(iii). We prove the following lemma, which is interesting in its
own right.
Lemma 8.5.2 (Lebesgue number lemma) Let M be a compact metric space and {Uα} an
open cover. Then there exists a real number δ > 0, called a Lebesgue number for that
covering, such that the δ-neighborhood of any point p lies entirely in some Uα.
Proof of lemma. Assume for contradiction that for every δ = 1∕k there is a point
xk∈ M such that its 1∕k-neighborhood isn’t contained in any Uα. In this way we
construct a sequence x1, x2, …; thus we’re allowed to take a subsequence which converges
to some x. Then for every 𝜀 > 0 we can find an integer n such that d(xn,x)+1∕n < 𝜀; thus
the 𝜀-neighborhood at x isn’t contained in any Uα for every 𝜀 > 0. This is impossible,
because we assumed x was covered by some open set. ■
Now, take a Lebesgue number δ for the covering. Since M is totally bounded, finitely many
δ-neighborhoods cover the space, so finitely many Uα do as well. □
Proof that (iii)(ii). One step is immediate:
Question 8.5.3.Show that the covering conditiontotally bounded.
The tricky part is showing M is complete. Assume for contradiction it isn’t and
thus that the sequence (xk) is Cauchy, but it doesn’t converge to any particular
point.
Question 8.5.4.Show that this implies for each p ∈ M, there is an 𝜀p-neighborhood Upwhichcontains at most finitely many of the points of the sequence (xk). (You will have to use the factthat xk↛p and (xk) is Cauchy.)
Now if we consider M = ⋃pUp we get a finite subcover of these open neighborhoods; but
this finite subcover can only cover finitely many points of the sequence, by contradiction. □
8.6A few harder problems to think about
The later problems are pretty hard; some have the flavor of IMO 3/6-style constructions. It’s
important to draw lots of pictures so one can tell what’s happening. Of these ?? is definitely
my favorite.
Problem 8A.Show that the closed interval [0,1] and open interval (0,1) are not
homeomorphic.
Problem 8B.Let X be a topological space with the discrete topology. Under what
conditions is X compact?
Problem 8C (The cofinite topology is quasicompact only).We let X be an infinite
set and equip it with the cofinite topology: the open sets are the empty set and
complements of finite sets. This makes X into a topological space. Show that X is
quasicompact but not Hausdorff.
Problem 8D (Cantor’s intersection theorem). Let X be a compact topological space,
and suppose
is an infinite sequence of nested nonempty closed subsets. Show that ⋂n≥0Kn≠∅.
Problem 8E (Tychonoff’s theorem).Let X and Y be compact metric spaces. Show
that X × Y is compact. (This is also true for general topological spaces, but the
proof is surprisingly hard, and we haven’t even defined X × Y in general yet.)
Problem 8F† (Bolzano-Weierstraß theorem for general metric spaces).Prove
that a metric space M is sequentially compact if and only if it is complete and totally
bounded.
Problem 8G (Almost Arzelà-Ascoli theorem).Let f1,f2,…: [0,1] → [−100,100] be an equicontinuous sequence of
functions, meaning
Show that we can extract a subsequence fi1,fi2,… of these functions such that for
every x ∈ [0,1], the sequence fi1(x), fi2(x), … converges.
Problem 8H.Let M = (M,d) be a bounded metric space. Suppose
that whenever d′ is another metric on M for which (M,d) and (M,d′) are homeomorphic
(i.e. have the same open sets), then d′ is also bounded. Prove that M is compact.
Problem 8I.In this problem a “circle” refers to the boundary of a disk
with nonzero radius.
(a)
Is it possible to partition the plane ℝ2 into disjoint circles?
(b)
From the plane ℝ2 we delete two distinct points p and q. Is it possible to partition
the remaining points into disjoint circles?
Part IV Linear Algebra
9Vector spaces
This is a pretty light chapter. The point of it is to define what a vector space and a basis are.
These are intuitive concepts that you may already know.
9.1The definitions of a ring and field
Prototypical example for this section:ℤ, ℝ, and ℂ are rings; the latter two are fields.
I’ll very informally define a ring/field here, in case you skipped the earlier chapter.
A ring is a structure with a commutative addition and multiplication, as well as
subtraction, like ℤ. It also has an additive identity 0 and multiplicative identity 1.
If the multiplication is invertible like in ℝ or ℂ, (meaning makes sense for any
x≠0), then the ring is called a field.
In fact, if you replace “field” by “ℝ” everywhere in what follows, you probably won’t lose much.
It’s customary to use the letter R for rings, and k or K for fields.
Finally, in case you skipped the chapter on groups, I should also mention:
An additive abelian group is a structure with a commutative addition, as well
as subtraction, plus an additive identity 0. It doesn’t have to have multiplication.
A good example is ℝ3 (with addition componentwise).
9.2Modules and vector spaces
Prototypical example for this section:Polynomials of degree at most n.
You intuitively know already that ℝn is a “vector space”: its elements can be
added together, and there’s some scaling by real numbers. Let’s develop this more
generally.
Fix a commutative ring R. Then informally,
An R-module is any structure where you can add two elements and scale byelements of R.
Moreover, a vector space is just a module whose commanding ring is actually a field. I’ll
give you the full definition in a moment, but first, examples…
Example 9.2.1 (Quadratic polynomials, aka my favorite example) My favorite
example of an ℝ-vector space is the set of polynomials of degree at most two, namely
Indeed, you can add any two quadratics, and multiply by constants. You can’t multiply
two quadratics to get a quadratic, but that’s irrelevant – in a vector space there need
not be a notion of multiplying two vectors together.
In a sense we’ll define later, this vector space has dimension 3 (as expected!).
Example 9.2.2 (All polynomials) The set of all polynomials with real coefficients is an
ℝ-vector space, because you can add any two polynomials and scale by constants.
Example 9.2.3 (Euclidean space)
(a)
The complex numbers
form a real vector space. As we’ll see later, it has “dimension 2”.
(b)
The real numbers ℝ form a real vector space of dimension 1.
(c)
The set of 3D vectors
forms a real vector space, because you can add any two triples component-wise.
Again, we’ll later explain why it has “dimension 3”.
Example 9.2.4 (More examples of vector spaces)
(a)
The set
has a structure of a ℚ-vector space in the obvious fashion: one can add any two
elements, and scale by rational numbers. (It is not a real vector space – why?)
(b)
The set
is a 2-dimensional real vector space.
(c)
The set of all functions f : ℝ → ℝ is also a real vector space (since the notions f +g
and c ⋅ f both make sense for c ∈ ℝ).
Now let me write the actual rules for how this multiplication behaves.
Definition 9.2.5.Let R be a commutative ring. An R-module starts with an additive
abelian group M = (M,+) whose identity is denoted 0 = 0M. We additionally specify a left
multiplication by elements of R. This multiplication must satisfy the following properties for
r,r1,r2∈ R and m,m1,m2∈ M:
(i)
r1⋅ (r2⋅ m) = (r1r2) ⋅ m.
(ii)
Multiplication is distributive, meaning
(iii)
1R⋅ m = m.
(iv)
0R⋅ m = 0M. (This is actually extraneous; one can deduce it from the first three.)
If R is a field we say M is an R-vector space; its elements are called vectors and the members
of R are called scalars.
Abuse of Notation 9.2.6.In the above, we’re using the same symbol + for the
addition of M and the addition of R. Sorry about that, but it’s kind of hard to avoid,
and the point of the axioms is that these additions should be related. I’ll try to remember
to put r ⋅ m for the multiplication of the module and r1r2 for the multiplication of R.
Question 9.2.7.In?? , I was careful to say “degree at most 2” instead of “degree 2”. What’sthe reason for this? In other words, why is
not an ℝ-vector space?
A couple less intuitive but somewhat important examples…
Example 9.2.8 (Abelian groups are ℤ-modules) (Skip this example if you’re not comfortable
with groups.)
(a)
The example of real polynomials
is also a ℤ-module! Indeed, we can add any two such polynomials, and we can scale
them by integers.
(b)
The set of integers modulo 100, say ℤ∕100ℤ, is a ℤ-module as well. Can you see
how?
(c)
In fact, any abelian group G = (G,+) is a ℤ-module. The multiplication can be
defined by
for n ≥ 0. (Here −g is the additive inverse of g.)
Example 9.2.9 (Every ring is its own module)
(a)
ℝ can be thought of as an ℝ-vector space over itself. Can you see why?
(b)
By the same reasoning, we see that any commutative ring R can be thought of as
an R-module over itself.
9.3Direct sums
Prototypical example for this section:{ax2 + bx + c} = ℝ ⊕xℝ ⊕x2ℝ, and ℝ3is the sum of itsaxes.
Let’s return to ?? , and consider
Even though I haven’t told you what a dimension is, you can probably see that this vector
space “should have” dimension 3. We’ll get to that in a moment.
The other thing you may have noticed is that somehow the x2, x and 1 terms don’t “talk to
each other”. They’re totally unrelated. In other words, we can consider the three
sets
x2ℝ
:=
xℝ
: =
ℝ
: = .
In an obvious way, each of these can be thought of as a “copy” of ℝ.
Then V quite literally consists of the “sums of these sets”. Specifically, every element of V
can be written uniquely as the sum of one element from each of these sets. This motivates us
to write
The notion which captures this formally is the direct sum.
Definition 9.3.1.Let M be an R-module. Let M1 and M2 be subsets of M which are
themselves R-modules. Then we write M = M1⊕ M2 and say M is a direct sum of
M1 and M2 if every element from M can be written uniquely as the sum of an element
from M1 and M2.
Example 9.3.2 (Euclidean plane) Take the vector space ℝ2 = .
We can consider it as a direct sum of its x-axis and y-axis:
Then ℝ2 = X ⊕ Y .
This gives us a “top-down” way to break down modules into some disconnected
components.
By applying this idea in reverse, we can also construct new vector spaces as follows. In a
very unfortunate accident, the two names and notations for technically distinct things are
exactly the same.
Definition 9.3.3.Let M and N be R-modules. We define the direct sumM ⊕ N to
be the R-module whose elements are pairs (m,n) ∈ M × N. The operations are given
by
and
For example, while we technically wrote ℝ2 = X ⊕Y , since each of X and Y is a copy of ℝ,
we might as well have written ℝ2ℝ ⊕ ℝ.
Abuse of Notation 9.3.4.The above illustrates an abuse of notation in the way we write a
direct sum. The symbol ⊕ has two meanings.
If V is a given space and W1 and W2 are subspaces, then V = W1⊕ W2 means
that “V splits as a direct sum W1⊕ W2” in the way we defined above.
If W1 and W2 are two unrelated spaces, then W1⊕ W2 is defined as the vector
space whose elements are pairs (w1,w2) ∈ W1× W2.
You can see that these definitions “kind of” coincide.
In this way, you can see that V should be isomorphic to ℝ ⊕ ℝ ⊕ ℝ; we had V = x2ℝ ⊕xℝ ⊕ ℝ,
but the 1, x, x2 don’t really talk to each other and each of the summands is really just a copy
of ℝ at heart.
Definition 9.3.5.We can also define, for every positive integer n, the module
9.4Linear independence, spans, and basis
Prototypical example for this section:is a basis of .
The idea of a basis, the topic of this section, gives us another way to capture the notion
that
is sums of copies of {1,x,x2}. This section should be very intuitive, if technical. If you can’t
see why the theorems here “should” be true, you’re doing it wrong.
Let M be an R-module now. We define three very classical notions that you likely are
already familiar with. If not, fall upon your notion of Euclidean space or V above.
Definition 9.4.1.A linear combination of some vectors v1,…,vn is a sum of the
form r1v1 + + rnvn, where r1,…,rn∈ R. The linear combination is called trivial if
r1 = r2 = = rn = 0R, and nontrivial otherwise.
Definition 9.4.2.Consider a finite set of vectors v1,…,vn in a module M.
It is called linearly independent if there is no nontrivial linear combination with
value 0M. (Observe that 0M = 0 ⋅ v1 + 0 ⋅ v2 + + 0 ⋅ vn is always true – the
assertion is that there is no other way to express 0M in this form.)
It is called a generating set if every v ∈ M can be written as a linear combination
of the {vi}. If M is a vector space we say it is spanning instead.
It is called a basis (plural bases) if every v ∈ M can be written uniquely as a
linear combination of the {vi}.
The same definitions apply for an infinite set, with the proviso that all sums must be
finite.
So by definition, is a basis for V . It’s not the only one: {2,x,x2} and
{x + 4,x − 2,x2 + x} are other examples of bases, though not as natural. However, the set
S = {3 + x2,x + 1,5 + 2x + x2} is not a basis; it fails for two reasons:
Note that 0 = (3 + x2) + 2(x + 1) − (5 + 2x + x2). So the set S is not linearly
independent.
It’s not possible to write x2 as a sum of elements of S. So S fails to be spanning.
With these new terms, we can say a basis is a linearly independent and spanning
set.
Example 9.4.3 (More examples of bases)
(a)
Regard ℚ[] = as a ℚ-vector space. Then {1,} is a basis.
(b)
If V is the set of all real polynomials, there is an infinite basis {1,x,x2,…}. The
condition that we only use finitely many terms just says that the polynomials must
have finite degree (which is good).
(c)
Let V = {(x,y,z)∣x + y + z = 0 and x,y,z ∈ ℝ}. Then we expect there to be a basis of
size 2, but unlike previous examples there is no immediately “obvious” choice. Some
working examples include:
(1,−1,0) and (1,0,−1),
(0,1,−1) and (1,0,−1),
(5,3,−8) and (2,−1,−1).
Exercise 9.4.4.Show that a set of vectors is a basis if and only if it is linearly independent andspanning. (Think about the polynomial example if you get stuck.)
Now we state a few results which assert that bases in vector spaces behave as nicely as
possible.
Theorem 9.4.5 (Maximality and minimality of bases) Let V be a vector space over some field k
and take e1,…,en∈ V . The following are equivalent:
(a)
The ei form a basis.
(b)
The ei are spanning, but no proper subset is spanning.
(c)
The ei are linearly independent, but adding any other element of V makes them not
linearly independent.
Remark 9.4.6 — If we replace V by a general module M over a commutative ring R,
then (a) (b) and (a) (c) but not conversely.
Proof.Straightforward, do it yourself if you like. The key point to notice is that you
need to divide by scalars for the converse direction, hence V is required to be a vector
space instead of just a module for the implications (b) (a) and (c) (a). □
Theorem 9.4.7 (Dimension theorem for vector spaces) If a vector space V has a finite
basis, then every other basis has the same number of elements.
Proof.We prove something stronger: Assume v1,…,vn is a spanning set while w1,…,wm is
linearly independent. We claim that n ≥ m.
Question 9.4.8.Show that this claim is enough to imply the theorem.
Let A0 = {v1,…,vn} be the spanning set. Throw in w1: by the spanning condition,
w1 = c1v1 + + cnvn. There’s some nonzero coefficient, say cn. Thus
Thus A1 = {v1,…,vn−1,w1} is spanning. Now do the same thing, throwing in w2, and
deleting some element of the vi as before to get A2; the condition that the wi are linearly
independent ensures that some vi coefficient must always not be zero. Since we can eventually
get to Am, we have n ≥ m. □
Remark 9.4.9 (Generalizations)
The theorem is true for an infinite basis as well if we interpret “the number of
elements” as “cardinality”. This is confusing on a first read through, so we won’t
elaborate.
In fact, this is true for modules over any commutative ring. Interestingly, the proof
for the general case proceeds by reducing to the case of a vector space.
The dimension theorem, true to its name, lets us define the dimension of a vector space as
the size of any finite basis, if one exists. When it does exist we say V is finite-dimensional.
So for example,
has dimension three, because is a basis. That’s not the only basis: we could as
well have written
and gotten the exact same vector space. But the beauty of the theorem is that no matter
how we try to contrive the generating set, we always will get exactly three elements. That’s
why it makes sense to say V has dimension three.
On the other hand, the set of all polynomials ℝ[x] is infinite-dimensional (which should be
intuitively clear).
A basis e1,…,en of V is really cool because it means that to specify v ∈ V , I only have to
specify a1,…,an∈ k, and then let v = a1e1 + + anen. You can even think of v as
. To put it another way, if V is a k-vector space we always have
9.5Linear maps
Prototypical example for this section:Evaluation of {ax2 + bx + c}at x = 3.
We’ve seen homomorphisms and continuous maps. Now we’re about to see linear maps, the
structure preserving maps between vector spaces. Can you guess the definition?
Definition 9.5.1.Let V and W be vector spaces over the same field k. A linear map is a
map T : V → W such that:
(i)
We have T(v1+v2) = T(v1)+T(v2) for any v1,v2∈ V .1
(ii)
For any a ∈ k and v ∈ V , T(a ⋅ v) = a ⋅ T(v).
If this map is a bijection (equivalently, if it has an inverse), it is an isomorphism. We then say
V and W are isomorphic vector spaces and write VW.
Example 9.5.2 (Examples of linear maps)
(a)
For any vector spaces V and W there is a trivial linear map sending everything to
0W∈ W.
(b)
For any vector space V , there is the identity isomorphism id : V → V .
(c)
The map ℝ3→ ℝ by (a,b,c)4a + 2b + c is a linear map.
(d)
Let V be the set of real polynomials of degree at most 2. The map ℝ3→ V by
(a,b,c)ax2 + bx + c is an isomorphism.
(e)
Let V be the set of real polynomials of degree at most 2. The map V → ℝ by
ax2 + bx + c9a + 3b + c is a linear map, which can be described as “evaluation at
3”.
(f)
Let W be the set of functions ℝ → ℝ. The evaluation map W → ℝ by ff(0) is a
linear map.
(g)
There is a map of ℚ-vector spaces ℚ[] → ℚ[] called “multiply by ”; this
map sends a + b2b + a. This map is an isomorphism, because it has an
inverse “multiply by 1∕”.
In the expression T(a⋅v) = a⋅T(v), note that the first ⋅ is the multiplication of V and the
second ⋅ is the multiplication of W. Note that this notion of isomorphism really only cares
about the size of the basis:
Proposition 9.5.3 (n-dimensional vector spaces are isomorphic) If V is an n-dimensional
vector space, then Vk⊕n.
Question 9.5.4.Let e1, …, enbe a basis for V . What is the isomorphism? (Your first guess isprobably right.)
Remark 9.5.5 — You could technically say that all finite-dimensional vector spaces are
just k⊕n and that no other space is worth caring about. But this seems kind of rude.
Spaces often are more than just triples: ax2 + bx + c is a polynomial, and so it has some
“essence” to it that you’d lose if you compressed it into (a,b,c).
Moreover, a lot of spaces, like the set of vectors (x,y,z) with x + y + z = 0, do not have
an obvious choice of basis. Thus to cast such a space into k⊕n would require you to make
arbitrary decisions.
9.6What is a matrix?
Now I get to tell you what a matrix is: it’s a way of writing a linear map in terms of
bases.
Suppose we have a finite-dimensional vector space V with basis e1,…,em and a vector space
W with basis w1,…,wn. I also have a map T : V → W and I want to tell you what T is. It
would be awfully inconsiderate of me to try and tell you what T(v) is at every point v. In fact,
I only have to tell you what T(e1), …, T(em) are, because from there you can work out
T(a1e1 + + amem) for yourself:
Since the ei are a basis, that tells you all you need to know about T.
Example 9.6.1 (Extending linear maps) Let V = . Then
T(ax2 + bx + c) = aT(x2) + bT(x) + cT(1).
Now I can even be more concrete. I could tell you what T(e1) is, but seeing as I have a basis
of W, I can actually just tell you what T(e1) is in terms of this basis. Specifically, there are
unique a11,a21,…,an1∈ k such that
So rather than telling you the value of T(e1) in some abstract space W, I could
just tell you what a11,a21,…,an1 were. Then I’d repeat this for T(e2), T(e3), all
the way up to T(em), and that would tell you everything you need to know about
T.
That’s where the matrix T comes from! It’s a concise way of writing down all
mn numbers I need to tell you. To be explicit, the matrix for T is defined as the
array
T
= m columnsn rows
= .
Example 9.6.2 (An example of a matrix) Here is a concrete example in terms of a basis. Let
V = ℝ3 with basis e1, e2, e3 and let W = ℝ2 with basis w1, w2. If I have T : V → W then
uniquely determined by three values, for example:
T(e1)
= 4w1 + 7w2
T(e2)
= 2w1 + 3w2
T(e3)
= w1
The columns then correspond to T(e1), T(e2), T(e3):
Example 9.6.3 (An example of a matrix after choosing a basis) We again let V =
be the vector space of polynomials of degree at most 2. We fix the basis
1, x, x2 for it.
Consider the “evaluation at 3” map, a map V → ℝ. We pick 1 as the basis element of
the RHS; then we can write it as a 1 × 3 matrix
with the columns corresponding to T(1), T(x), T(x2).
From here you can actually work out for yourself what it means to multiply two matrices.
Suppose we have picked a basis for three spaces U, V , W. Given maps T : U → V and
S : V → W, we can consider their composition S ∘ T, i.e.
Matrix multiplication is defined exactly so that the matrix ST is the same thing we get from
interpreting the composed function S ∘ T as a matrix.
Exercise 9.6.4.Check this for yourself! For a concrete example let ℝ2ℝ2ℝ2by T(e1) =2e1+ 3e2and T(e2) = 4e1+ 5e2, S(e1) = 6e1+ 7e2and S(e2) = 8e1+ 9e2. Compute S(T(e1))and S(T(e2)) and see how it compares to multiplying the matrices associated to S and T.
In particular, since function composition is associative, it follows that matrix multiplication
is as well. To drive this point home,
A matrix is the laziest possible way to specify a linear map from V toW.
This means you can define concepts like the determinant or the trace of a matrix both in
terms of an “intrinsic” map T : V → W and in terms of the entries of the matrix. Since the
map T itself doesn’t refer to any basis, the abstract definition will imply that the numerical
definition doesn’t depend on the choice of a basis.
9.7Subspaces and picking convenient bases
Prototypical example for this section:Any two linearly independent vectors in ℝ3.
Definition 9.7.1.Let M be a left R-module. A submoduleN of M is a module N
such that every element of N is also an element of M. If M is a vector space then N is
called a subspace.
Example 9.7.2 (Kernels) The kernel of a map T : V → W (written kerT) is the set of
v ∈ V such that T(v) = 0W. It is a subspace of V , since it’s closed under addition and
scaling (why?).
Example 9.7.3 (Spans) Let V be a vector space and v1,…,vm be any vectors of V . The
span of these vectors is defined as the set
Note that it is a subspace of V as well!
Question 9.7.4.Why is 0Van element of each of the above examples? In general, why mustany subspace contain 0V?
Subspaces behave nicely with respect to bases.
Theorem 9.7.5 (Basis completion) Let V be an n-dimensional space, and V ′ a subspace of V .
Then
(a)
V ′ is also finite-dimensional.
(b)
If e1,…,em is a basis of V ′, then there exist em+1,…,en in V such that e1,…,en is a
basis of V .
Proof.Omitted, since it is intuitive and the proof is not that enlightening. (However,
we will use this result repeatedly later on, so do take the time to internalize it now.) □
A very common use case is picking a convenient basis for a map T.
Theorem 9.7.6 (Picking a basis for linear maps) Let T : V → W be a map of
finite-dimensional vector spaces, with n = dimV , m = dimW. Then there exists a basis
v1,…,vn of V and a basis w1,…,wm of W, as well as a nonnegative integer k, such that
Moreover dimkerT = n − k and dimTimg(V ) = k.
Sketch of Proof. You might like to try this one yourself before reading on: it’s a repeated
application of ?? .
Let kerT have dimension n−k. We can pick vk+1,…,vn a basis of kerT. Then extend
it to a basis v1,…,vn of V . The map T is injective over the span of v1,…,vk (since only
0V is in the kernel) so its images in W are linearly independent. Setting wi = T(vi) for
each i, we get some linearly independent set in W. Then extend it again to a basis of
W. □
This theorem is super important, not only because of applications but also because it will
give you the right picture in your head of how a linear map is supposed to look. I’ll even draw
a cartoon of it to make sure you remember:
In particular, for T : V → W, one can write V = kerT ⊕V ′, so that T annihilates its kernel
while sending V ′ to an isomorphic copy in W.
A corollary of this (which you should have expected anyways) is the so called rank-nullity
theorem, which is the analog of the first isomorphism theorem.
Theorem 9.7.7 (Rank-nullity theorem) Let V and W be finite-dimensional vector spaces.
If T : V → W, then
Question 9.7.8.Conclude the rank-nullity theorem from?? .
9.8A cute application: Lagrange interpolation
Here’s a cute application2
of linear algebra to a theorem from high school.
Theorem 9.8.1 (Lagrange interpolation) Let x1,…,xn+1 be distinct real numbers and
y1,…,yn+1 any real numbers. Then there exists a unique polynomial P of degree at most
n such that
for every i.
When n = 1 for example, this loosely says there is a unique line joining two points.
Proof.The idea is to consider the vector space V of polynomials with degree at most n, as
well as the vector space W = ℝn+1.
Question 9.8.2.Check thatdimV = n + 1 =dimW. This is easiest to do if you pick a basisfor V , but you can then immediately forget about the basis once you finish this exercise.
Then consider the linear map T : V → W given by
This is indeed a linear map because, well, T(P + Q) = T(P) + T(Q) and T(cP) = cT(P). It
also happens to be injective: if P ∈ kerT, then P(x1) = = P(xn+1) = 0, but deg P ≤ n and
so P can only be the zero polynomial.
So T is an injective map between vector spaces of the same dimension. Thus it is actually a
bijection, which is exactly what we wanted. □
9.9(Digression) Arrays of numbers are evil
As I’ll stress repeatedly, a matrix represents a linear map between two vector spaces. Writing it
in the form of an m × n matrix is merely a very convenient way to see the map
concretely. But it obfuscates the fact that this map is, well, a map, not an array of
numbers.
If you took high school precalculus, you’ll see everything done in terms of matrices. To any
typical high school student, a matrix is an array of numbers. No one is sure what exactly these
numbers represent, but they’re told how to magically multiply these arrays to get more arrays.
They’re told that the matrix
is an “identity matrix”, because when you multiply by another matrix it doesn’t change.
Then they’re told that the determinant is some magical combination of these numbers formed
by this weird multiplication rule. No one knows what this determinant does, other than the
fact that det(AB) = detAdetB, and something about areas and row operations and Cramer’s
rule.
Then you go into linear algebra in college, and you do more magic with these arrays of
numbers. You’re told that two matrices T1 and T2 are similar if
for some invertible matrix S. You’re told that the trace of a matrix TrT is the
sum of the diagonal entries. Somehow this doesn’t change if you look at a similar
matrix, but you’re not sure why. Then you define the characteristic polynomial
as
Somehow this also doesn’t change if you take a similar matrix, but now you really don’t
know why. And then you have the Cayley-Hamilton theorem in all its black magic: pT(T) is
the zero map. Out of curiosity you Google the proof, and you find some ad-hoc procedure
which still leaves you with no idea why it’s true.
This is terrible. What’s so special about T2 = ST1S−1? Only if you know that the matrices
are linear maps does this make sense: T2 is just T1 rewritten with a different choice of
basis.
I really want to push the opposite view. Linear algebra is the study of linear maps, but it is
taught as the study of arrays of numbers, and no one knows what these numbers mean. And
for a good reason: the numbers are meaningless. They are a highly convenient way of encoding
the matrix, but they are not the main objects of study, any more than the dates of events are
the main objects of study in history.
The other huge downside is that people get the impression that the only (real) vector space
in existence is ℝ⊕n. As explained in ?? , while you can work this way if you’re a soulless
robot, it’s very unnatural for humans to do so.
When I took Math 55a as a freshman at Harvard, I got the exact opposite treatment: we did
all of linear algebra without writing down a single matrix. During all this time I was
quite confused. What’s wrong with a basis? I didn’t appreciate until later that this
approach was the morally correct way to treat the subject: it made it clear what was
happening.
Throughout the Napkin, I’ve tried to strike a balance between these two approaches, using
matrices when appropriate to illustrate the maps and to simplify proofs, but ultimately
writing theorems and definitions in their morally correct form. I hope that this has both the
advantage of giving the “right” definitions while being concrete enough to be digested. But I
would like to say for the record that, if I had to pick between the high school approach and the
55a approach, I would pick 55a in a heartbeat.
9.10A word on general modules
Prototypical example for this section:ℤ[] is a ℤ-module of rank two.
I focused mostly on vector spaces (aka modules over a field) in this chapter for simplicity, so
I want to make a few remarks about modules over a general commutative ring R before
concluding.
Firstly, recall that for general modules, we say “generating set” instead of “spanning set”.
Shrug.
The main issue with rings is that our key theorem ?? fails in spectacular ways. For
example, consider ℤ as a ℤ-module over itself. Then {2} is linearly independent, but it cannot
be extended to a basis. Similarly, {2,3} is spanning, but one cannot cut it down to a basis.
You can see why defining dimension is going to be difficult.
Nonetheless, there are still analogs of some of the definitions above.
Definition 9.10.1.An R-module M is called finitely generated if it has a finite
generating set.
Definition 9.10.2.An R-module M is called free if it has a basis. As said before, the
analogue of the dimension theorem holds, and we use the word rank to denote the size
of the basis. As before, there’s an isomorphism MR⊕n where n is the rank.
Example 9.10.3 (An example of a ℤ-module) The ℤ-module
has a basis {1,}, so we say it is a free ℤ-module of rank 2.
Abuse of Notation 9.10.4 (Notation for groups).Recall that an abelian group can
be viewed a ℤ-module (and in fact vice-versa!), so we can (and will) apply these words
to abelian groups. We’ll use the notation G⊕H for two abelian groups G and H for their
Cartesian product, emphasizing the fact that G and H are abelian. This will happen
when we study algebraic number theory and homology groups.
9.11A few harder problems to think about
General hint: ?? will be your best friend for many of these problems.
Problem 9A†.Let V and W be finite-dimensional vector spaces with nonzero dimension, and
consider linear maps T : V → W. Complete the following table by writing “sometimes”,
“always”, or “never” for each entry.
T injective
T surjective
T isomorphism
If dimV > dimW…
If dimV = dimW…
If dimV < dimW…
Problem 9B† (Equal dimension vector spaces are usually isomorphisms). Let V and W be
finite-dimensional vector spaces with dimV = dimW. Prove that for a map T : V → W, the
following are equivalent:
T is injective,
T is surjective,
T is bijective.
Problem 9C (Multiplication by ).Let V = ℚ[] = be a two-dimensional
ℚ-vector space, and fix the basis {1,} for it. Write down the 2 × 2 matrix with rational
coefficients that corresponds to multiplication by .
Problem 9D (Multivariable Lagrange interpolation).Let S ⊂ ℤ2 be a set of n lattice
points. Prove that there exists a nonzero two-variable polynomial p with real coefficients,
of degree at most , such that p(x,y) = 0 for every (x,y) ∈ S.
Problem 9E (Putnam 2003).Do there exist polynomials a(x), b(x), c(y), d(y) such
that
holds identically?
Problem 9F (TSTST 2014).Let P(x) and Q(x) be arbitrary
polynomials with real coefficients, and let d be the degree of P(x). Assume that P(x) is
not the zero polynomial. Prove that there exist polynomials A(x) and B(x) such
that
(i)
Both A and B have degree at most d∕2,
(ii)
At most one of A and B is the zero polynomial,
(iii)
P divides A + Q ⋅ B.
Problem 9G⋆ (Idempotents are projection maps). Let P : V → V be a linear map,
where V is a vector space (not necessarily finite-dimensional). Suppose P is idempotent,
meaning P(P(v)) = P(v) for each v ∈ V , or equivalently P is the identity on its image.
Prove that
Thus we can think of P as projection onto the subspace imP.
Problem 9H⋆.Let V be a finite dimensional vector space. Let
T : V → V be a linear map, and let Tn: V → V denote T applied n times. Prove that
there exists an integer N such that
10Eigen-things
This chapter will develop the theory of eigenvalues and eigenvectors, the so-called “Jordan
canonical form”. (Later on we will use it to define the characteristic polynomial.)
10.1Why you should care
We know that a square matrix T is really just a linear map from V to V . What’s the simplest
type of linear map? It would just be multiplication by some scalar λ, which would have
associated matrix (in any basis!)
That’s perhaps too simple, though. If we had a fixed basis e1,…,en then another very
“simple” operation would just be scaling each basis element ei by λi, i.e. a diagonal matrix
of the form
These maps are more general. Indeed, you can, for example, compute T100 in a
heartbeat: the map sends e1→ λ1100e1. (Try doing that with an arbitrary n × n
matrix.)
Of course, most linear maps are probably not that nice. Or are they?
Example 10.1.1 (Getting lucky) Let V be some two-dimensional vector space with e1
and e2 as basis elements. Let’s consider a map T : V → V by e12e1 and e2e1 +3e2,
which you can even write concretely as
This doesn’t look anywhere as nice until we realize we can rewrite it as
e1
2e1
e1 + e2
3(e1 + e2).
So suppose we change to the basis e1 and e1 + e2. Thus in the new basis,
So our completely random-looking map, under a suitable change of basis, looks like the very
nice maps we described before!
In this chapter, we will be making our luck, and we will see that our better understanding of
matrices gives us the right way to think about this.
10.2Warning on assumptions
Most theorems in this chapter only work for
finite-dimensional vector spaces V ,
over a field k which is algebraically closed.
On the other hand, the definitions work fine without these assumptions.
10.3Eigenvectors and eigenvalues
Let k be a field and V a vector space over it. In the above example, we saw that there were
two very nice vectors, e1 and e1 + e2, for which V did something very simple. Naturally, these
vectors have a name.
Definition 10.3.1.Let T : V → V and v ∈ V a nonzero vector. We say that v is an
eigenvector if T(v) = λv for some λ ∈ k (possibly zero, but remember v≠0). The value
λ is called an eigenvalue of T.
We will sometimes abbreviate “v is an eigenvector with eigenvalue λ” to just “v is a
λ-eigenvector”.
Of course, no mention to a basis anywhere.
Example 10.3.2 (An example of an eigenvector and eigenvalue) Consider the example earlier
with T = .
(a)
Note that e1 and e1 + e2 are 2-eigenvectors and 3-eigenvectors.
(b)
Of course, 5e1 is also an 2-eigenvector.
(c)
And, 7e1 + 7e2 is also a 3-eigenvector.
So you can quickly see the following observation.
Question 10.3.3.Show that the λ-eigenvectors, together with {0} form a subspace.
Definition 10.3.4.For any λ, we define the λ-eigenspace as the set of λ-eigenvectors
together with 0.
This lets us state succinctly that “2 is an eigenvalue of T with one-dimensional eigenspace
spanned by e1”.
Unfortunately, it’s not exactly true that eigenvalues always exist.
Example 10.3.5 (Eigenvalues need not exist) Let V = ℝ2 and let T be the map which
rotates a vector by 90∘ around the origin. Then T(v) is not a multiple of v for any v ∈ V ,
other than the trivial v = 0.
However, it is true if we replace k with an algebraically closed
field1 .
Theorem 10.3.6 (Eigenvalues always exist over algebraically closed fields) Suppose k is an
algebraically closed field. Let V be a finite dimensional k-vector space. Then if T : V → V
is a linear map, there exists an eigenvalue λ ∈ k.
Proof.(From [?]) The idea behind this proof is to consider “polynomials” in T. For example,
2T2− 4T + 5 would be shorthand for 2T(T(v)) − 4T(v) + 5v. In this way we can consider
“polynomials” P(T); this lets us tie in the “algebraically closed” condition. These polynomials
behave nicely:
Question 10.3.7.Show that P(T) + Q(T) = (P + Q)(T) and P(T) ∘ Q(T) = (P ⋅ Q)(T).
Let n = dimV < ∞ and fix any nonzero vector v ∈ V , and consider vectors v, T(v), …,
Tn(v). There are n + 1 of them, so they can’t be linearly independent for dimension reasons;
thus there is a nonzero polynomial P such that P(T) is zero when applied to v. WLOG
suppose P is a monic polynomial, and thus P(z) = (z − r1)…(z − rm) say. Then we
get
(where id is the identity matrix). This means at least one of T −riid is not injective, i.e. has
a nontrivial kernel, which is the same as an eigenvector. □
So in general we like to consider algebraically closed fields. This is not a big loss: any real
matrix can be interpreted as a complex matrix whose entries just happen to be real, for
example.
10.4The Jordan form
So that you know exactly where I’m going, here’s the main theorem.
Definition 10.4.1.A Jordan block is an n × n matrix of the following shape:
In other words, it has λ on the diagonal, and 1 above it. We allow n = 1, so is a
Jordan block.
Theorem 10.4.2 (Jordan canonical form) Let T : V → V be a linear map of
finite-dimensional vector spaces over an algebraically closed field k. Then we can choose
a basis of V such that the matrix T is “block-diagonal” with each block being a Jordan
block.
Such a matrix is said to be in Jordan form. This form is unique up to rearranging the
order of the blocks.
As an example, this means the matrix should look something like:
Question 10.4.3.Check that diagonal matrices are the special case when each block is 1 × 1.
What does this mean? Basically, it means our dream is almost true. What happens is that
V can get broken down as a direct sum
and T acts on each of these subspaces independently. These subspaces correspond to the
blocks in the matrix above. In the simplest case, dimJi = 1, so Ji has a basis element e for
which T(e) = λie; in other words, we just have a simple eigenvalue. But on occasion, the
situation is not quite so simple, and we have a block of size greater than 1; this leads to 1’s
just above the diagonals.
I’ll explain later how to interpret the 1’s, when I make up the word descending staircase. For
now, you should note that even if dimJi≥ 2, we still have a basis element which is an
eigenvector with eigenvalue λi.
Example 10.4.4 (A concrete example of Jordan form) Let T : k6→ k6 and suppose T
is given by the matrix
Reading the matrix, we can compute all the eigenvectors and eigenvalues: for any constants
a,b ∈ k we have
T(a ⋅ e1)
= 5a ⋅ e1
T(a ⋅ e2)
= 2a ⋅ e2
T(a ⋅ e4)
= 7a ⋅ e4
T(a ⋅ e5 + b ⋅ e6)
= 3.
The element e3 on the other hand, is not an eigenvector since T(e3) = e2 + 2e3.
10.5Nilpotent maps
Bear with me for a moment. First, define:
Definition 10.5.1.A map T : V → V is nilpotent if Tm is the zero map for some
integer m. (Here Tm means “T applied m times”.)
What’s an example of a nilpotent map?
Example 10.5.2 (The “descending staircase”) Let V = k⊕3 have basis e1, e2, e3. Then
the map T which sends
is nilpotent, since T(e1) = T2(e2) = T3(e3) = 0, and hence T3(v) = 0 for all v ∈ V .
The 3 × 3 descending staircase has matrix representation
You’ll notice this is a Jordan block.
Exercise 10.5.3.Show that the descending staircase above has 0 as its only eigenvalue.
That’s a pretty nice example. As another example, we can have multiple such
staircases.
Example 10.5.4 (Double staircase) Let V = k⊕5 have basis e1, e2, e3, e4, e5. Then the
map
is nilpotent.
Picture, with some zeros omitted for emphasis:
You can see this isn’t really that different from the previous example; it’s just the same idea
repeated multiple times. And in fact we now claim that all nilpotent maps have essentially
that form.
Theorem 10.5.5 (Nilpotent Jordan) Let V be a finite-dimensional vector space over an
algebraically closed field k. Let T : V → V be a nilpotent map. Then we can write
V = ⊕i=1mVi where each Vi has a basis of the form vi, T(vi), …, TdimVi−1(vi) for some
vi∈ Vi.
Hence:
Every nilpotent map can be viewed as independent staircases.
Each chain vi, T(vi), T(T(vi)), …is just one staircase. The proof is given later, but first let
me point out where this is going.
Here’s the punch line. Let’s take the double staircase again. Expressing it as a matrix gives,
say
Then we can compute
It’s a bunch of λ Jordan blocks! This gives us a plan to proceed: we need to break V into a
bunch of subspaces such that T −λid is nilpotent over each subspace. Then Nilpotent Jordan
will finish the job.
10.6Reducing to the nilpotent case
Definition 10.6.1.Let T : V → V . A subspace W ⊆ V is called T-invariant if
T(w) ∈ W for any w ∈ W. In this way, T can be thought of as a map W → W.
In this way, the Jordan form is a decomposition of V into invariant subspaces.
Now I’m going to be cheap, and define:
Definition 10.6.2.A map T : V → V is called indecomposable if it’s impossible to
write V = W1⊕ W2 where both W1 and W2 are nontrivial T-invariant spaces.
Picture of a decomposable map:
As you might expect, we can break a space apart into “indecomposable” parts.
Proposition 10.6.3 (Invariant subspace decomposition) Let V be a finite-dimensional
vector space. Given any map T : V → V , we can write
where each Vi is T-invariant, and for any i the map T : Vi→ Vi is indecomposable.
Proof.Same as the proof that every integer is the product of primes. If V is not
decomposable, we are done. Otherwise, by definition write V = W1⊕ W2 and then
repeat on each of W1 and W2. □
Incredibly, with just that we’re almost done! Consider a decomposition as above, so that
T : V1→ V1 is an indecomposable map. Then T has an eigenvalue λ1, so let S = T − λ1id;
hence kerS≠{0}.
Question 10.6.4.Show that V1is also S-invariant, so we can consider S : V1→ V1.
By ?? , we have
for some N. But we assumed T was indecomposable, so this can only happen if imSN = {0}
and kerSN = V1 (since kerSN contains our eigenvector). Hence S is nilpotent, so it’s a
collection of staircases. In fact, since T is indecomposable, there is only one staircase. Hence
V1 is a Jordan block, as desired.
10.7(Optional) Proof of nilpotent Jordan
The proof is just induction on dimV . Assume dimV ≥ 1, and let W = Timg(V ) be the image
of V . Since T is nilpotent, we must have W ⊊ V . Moreover, if W = {0} (i.e. T is the zero
map) then we’re already done. So assume {0} ⊊ W ⊊ V .
By the inductive hypothesis, we can select a good basis of W:
ℬ′ =
T(v1),T(T(v1)),T(T(T(v1))),…
T(v2),T(T(v2)),T(T(T(v2))),…
…,
T(vℓ),T(T(vℓ)),T(T(T(vℓ))),…
for some T(vi) ∈ W (here we have taken advantage of the fact that each element of W is
itself of the form T(v) for some v).
Also, note that there are exactly ℓ elements of ℬ′ which are in kerT (namely the last
element of each of the ℓ staircases). We can thus complete it to a basis vℓ+1,…,vm (where
m = dimkerT). (In other words, the last element of each staircase plus the m−ℓ new ones are
a basis for kerT.)
Now consider
ℬ =
v1,T(v1),T(T(v1)),T(T(T(v1))),…
v2,T(v2),T(T(v2)),T(T(T(v2))),…
…,
vℓ,T(vℓ),T(T(vℓ)),T(T(T(vℓ))),…
vℓ+1,vℓ+2,…,vm.
Question 10.7.1.Check that there are exactly ℓ +dimW + (dimkerT − ℓ) =dimV elements.
Exercise 10.7.2.Show that all the elements are linearly independent. (Assume for contradictionthere is some linear dependence, then take T of both sides.)
Hence ℬ is a basis of the desired form.
10.8Algebraic and geometric multiplicity
Prototypical example for this section:The matrix T below.
This is some convenient notation: let’s consider the matrix in Jordan form
We focus on the eigenvalue 7, which appears multiple times, so it is certainly
“repeated”. However, there a two different senses in which you could say it is repeated.
Algebraic: You could say it is repeated five times, because it appears five times on
the diagonal.
Geometric: You could say it really only appears two times: because there are only
two eigenvectors with eigenvalue 7, namely e1 and e4.
Indeed, the vector e2 for example has T(e2) = 7e2 + e1, so it’s not really an
eigenvector! If you apply T − 7id to e2 twice though, you do get zero.
Question 10.8.1.In this example, how many times do you need to apply T − 7idto e6to getzero?
Both these notions are valid, so we will name both. To preserve generality, we first state the
“intrinsic” definition.
Definition 10.8.2.Let T : V → V be a linear map and λ a scalar.
The geometric multiplicity of λ is the dimension dimVλ of the λ-eigenspace.
Define the generalized eigenspaceVλ to be the subspace of V for which
(T − λid)n(v) = 0 for some n ≥ 1. The algebraic multiplicity of λ is the
dimension dimVλ.
(Silly edge case: we allow “multiplicity zero” if λ is not an eigenvalue at all.)
However in practice you should just count the Jordan blocks.
Example 10.8.3 (An example of eigenspaces via Jordan form) Retain the matrix T
mentioned earlier and let λ = 7.
The eigenspace Vλ has basis e1 and e4, so the geometric multiplicity is 2.
The generalized eigenspace Vλ has basis e1, e2, e4, e5, e6 so the algebraic
multiplicity is 5.
To be completely explicit, here is how you think of these in practice:
Proposition 10.8.4 (Geometric and algebraic multiplicity vs Jordan blocks) Assume T : V → V is
a linear map of finite-dimensional vector spaces, written in Jordan form. Let λ be a scalar.
Then
The geometric multiplicity of λ is the number of Jordan blocks with eigenvalue λ;
the eigenspace has one basis element per Jordan block.
The algebraic multiplicity of λ is the sum of the dimensions of the Jordan blocks
with eigenvalue λ; the eigenspace is the direct sum of the subspaces corresponding
to those blocks.
Question 10.8.5.Show that the geometric multiplicity is always less than or equal to thealgebraic multiplicity.
This actually gives us a tentative definition:
The trace is the sum of the eigenvalues, counted with algebraic multiplicity.
The determinant is the product of the eigenvalues, counted with algebraic
multiplicity.
This definition is okay, but it has the disadvantage of requiring the ground field to be
algebraically closed. It is also not the definition that is easiest to work with computationally.
The next two chapters will give us a better definition.
10.9A few harder problems to think about
Problem 10A (Sum of algebraic multiplicities).Given a 2018-dimensional complex vector
space V and a map T : V → V , what is the sum of the algebraic multiplicities of all
eigenvalues of T?
Problem 10B (The word “diagonalizable”).A linear map T : V → V (where dimV is
finite) is said to be diagonalizable if it has a basis e1, …, en such that each ei is an
eigenvector.
(a)
Explain the name “diagonalizable”.
(b)
Suppose we are working over an algebraically closed field. Then show that that T
is diagonalizable if and only if for any λ, the geometric multiplicity of λ equals the
algebraic multiplicity of λ.
Problem 10C (Switcharoo).Let V be the ℂ-vector space with basis e1 and e2. The
map T : V → V sends T(e1) = e2 and T(e2) = e1. Determine the eigenspaces of T.
Problem 10D (Writing a polynomial backwards).Define the complex vector space V
of polynomials with degree at most 2, say V = . Define
T : V → V by
Determine the eigenspaces of T.
Problem 10E (Differentiation of polynomials).Let V = ℝ[x] be the real vector space of all
real polynomials. Note that : V → V is a linear map (for example it sends x3 to 3x2).
Which real numbers are eigenvalues of this map?
Problem 10F (Differentiation of functions).Let V be the real vector space of all
infinitely differentiable functions ℝ → ℝ. Note that : V → V is a linear map (for
example it sends cosx to −sinx). Which real numbers are eigenvalues of this map?
11Dual space and trace
You may have learned in high school that given a matrix
the trace is the sum along the diagonals a + d and the determinant is ad − bc. But
we know that a matrix is somehow just encoding a linear map using a choice of
basis. Why would these random formulas somehow not depend on the choice of a
basis?
In this chapter, we are going to give an intrinsic definition of TrT, where T : V → V and
dimV < ∞. This will give a coordinate-free definition which will in particular imply the trace
a + d doesn’t change if we take a different basis.
In doing so, we will introduce two new constructions: the tensor product V ⊗W (which is a
sort of product of two spaces, with dimension dimV ⋅ dimW) and the dual space V∨, which is
the set of linear maps V → k (a k-vector space). Later on, when we upgrade from a vector
space V to an inner product space V∨, we will see that the dual space gives a nice
interpretation of the “transpose” of a matrix. You’ll already see some of that come through
here.
The trace is only defined for finite-dimensional vector spaces, so if you want you can restrict
your attention to finite-dimensional vector spaces for this chapter. (On the other hand we do
not need the ground field to be algebraically closed.)
The next chapter will then do the same for the determinant.
11.1Tensor product
Prototypical example for this section:ℝ[x] ⊗ ℝ[y] = ℝ[x,y].
We know that dim(V ⊕ W) = dimV + dimW, even though as sets V ⊕ W looks
like V × W. What if we wanted a real “product” of spaces, with multiplication of
dimensions?
For example, let’s pull out my favorite example of a real vector space, namely
Here’s another space, a little smaller:
If we take the direct sum, then we would get some rather unnatural vector space of
dimension five (whose elements can be thought of as pairs (ax2 + bx + c,dy + e)). But suppose
we want a vector space whose elements are products of polynomials in V and W;
it would contain elements like 4x2y + 5xy + y + 3. In particular, the basis would
be
and thus have dimension six.
For this we resort to the tensor product. It does exactly this, except that the “multiplication” is done
by a scary1
symbol ⊗: think of it as a “wall” that separates the elements between the two vector spaces.
For example, the above example might be written as
(This should be read as (4x2⊗y) + (5x⊗y) + …; addition comes after ⊗.) Of course there
should be no distinction between writing 4x2⊗ y and x2⊗ 4y or even 2x2⊗ 2y.
While we want to keep the x and y separate, the scalars should be free to float
around.
Of course, there’s no need to do everything in terms of just the monomials. We are free to
write
If you like, you can expand this as
Same thing. The point is that we can take any two of our polynomials and artificially
“tensor” them together.
The definition of the tensor product does exactly this, and nothing
else.2
Definition 11.1.1.Let V and W be vector spaces over the same field k. The tensorproductV ⊗kW is the abelian group generated by elements of the form v ⊗ w, subject to
relations
(v1 + v2) ⊗ w
= v1⊗ w + v2⊗ w
v ⊗ (w1 + w2)
= v ⊗ w1 + v ⊗ w2
(c ⋅ v) ⊗ w
= v ⊗ (c ⋅ w).
As a vector space, its action is given by c ⋅ (v ⊗ w) = (c ⋅ v) ⊗ w = v ⊗ (c ⋅ w).
Here’s another way to phrase the same idea. We define a pure tensor as an element of the
form v ⊗ w for v ∈ V and w ∈ W. But we let the ⊗ wall be “permeable” in the sense
that
and we let multiplication and addition distribute as we expect. Then V ⊗ W consists of
sums of pure tensors.
Example 11.1.2 (Infinite-dimensional example of tensor product: two-variable polynomials) Although it’s not relevant to this chapter, this definition works equally well with infinite-dimensional
vector spaces. The best example might be
That is, the tensor product of polynomials in x with real polynomials in y turns out to
just be two-variable polynomials ℝ[x,y].
Remark 11.1.3 (Warning on sums of pure tensors) — Remember the elements of V ⊗kW really are sums of these pure tensors! If you liked the previous example, this fact
has a nice interpretation — not every polynomial in ℝ[x,y] = ℝ[x] ⊗ℝℝ[y] factors as a
polynomial in x times a polynomial in y (i.e. as pure tensors f(x) ⊗ g(y)). But they all
can be written as sums of pure tensors xa⊗ yb.
As the example we gave suggested, the basis of V ⊗kW is literally the “product” of the bases of
V and W. In particular, this fulfills our desire that dim(V ⊗kW) = dimV ⋅ dimW.
Proposition 11.1.4 (Basis of V ⊗ W) Let V and W be finite-dimensional k-vector spaces.
If e1,…,em is a basis of V and f1,…,fn is a basis of W, then the basis of V ⊗kW is
precisely ei⊗ fj, where i = 1,…,m and j = 1,…,n.
Proof.Omitted; it’s easy at least to see that this basis is spanning. □
Example 11.1.5 (Explicit computation) Let V have basis e1, e2 and W have basis f1,f2.
Let v = 3e1 + 4e2∈ V and w = 5f1 + 6f2∈ W. Let’s write v ⊗ w in this basis for
V ⊗kW:
So you can see why tensor products are a nice “product” to consider if we’re really interested
in V × W in a way that’s more intimate than just a direct sum.
Abuse of Notation 11.1.6.Moving forward, we’ll almost always abbreviate ⊗k to just
⊗, since k is usually clear.
Remark 11.1.7 — Observe that to define a linear map V ⊗W → X, I only have to say
what happens to each pure tensor v ⊗ w, since the pure tensors generate V ⊗ W. But
again, keep in mind that V ⊗ W consists of sums of these pure tensors! In other words,
V ⊗ W is generated by pure tensors.
Remark 11.1.8 — Much like the Cartesian product A × B of sets, you can tensor
together any two vector spaces V and W over the same field k; the relationship between
V and W is completely irrelevant. One can think of the ⊗ as a “wall” through which one
can pass scalars in k, but otherwise keeps the elements of V and W separated. Thus, ⊗
is content-agnostic.
This also means that even if V and W have some relation to each other, the tensor
product doesn’t remember this. So for example v ⊗ 1≠1 ⊗ v, just like (g,1G)≠(1G,g) in
the group G × G.
11.2Dual space
Prototypical example for this section:Rotate a column matrix by 90 degrees.
Consider the following vector space:
Example 11.2.1 (Functions from ℝ3→ ℝ) The set of real functions f(x,y,z) is an
infinite-dimensional real vector space. Indeed, we can add two functions to get f + g,
and we can think of functions like 2f.
This is a terrifyingly large vector space, but you can do some reasonable reductions.
For example, you can restrict your attention to just the linear maps from ℝ3 to
ℝ.
That’s exactly what we’re about to do. This definition might seem strange at first, but bear
with me.
Definition 11.2.2.Let V be a k-vector space. Then V∨, the dual space of V , is defined
as the vector space whose elements are linear maps from V to k.
The addition and multiplication are pointwise: it’s the same notation we use when
we write cf + g to mean c ⋅ f(x) + g(x). The dual space itself is less easy to think
about.
Let’s try to find a basis for V∨. First, here is a very concrete interpretation of the vector
space. Suppose for example V = ℝ3. We can think of elements of V as column matrices,
like
Then a linear map f : V → k can be interpreted as a row matrix:
Then
More precisely: to specify a linear map V → k, I only have to tell you where eachbasis element of V goes. In the above example, f sends e1 to 3, e2 to 4, and e3 to 5. So f
sends
Let’s make all this precise.
Proposition 11.2.3 (The dual basis for V∨) Let V be a finite-dimensional vector space
with basis e1,…,en. For each i consider the function ei∨: V → k defined by
In more humane terms, ei∨(v) gives the coefficient of ei in v.
Then e1∨, e2∨, …, en∨ is a basis of V∨.
Example 11.2.4 (Explicit example of element in V∨) In this notation, f = 3e1∨ + 4e2∨ + 5e3∨.
Do you see why the “sum” notation works as expected here? Indeed
f(e1)
= (3e1∨ + 4e2∨ + 5e3∨)(e1)
= 3e1∨(e1) + 4e2∨(e1) + 5e3∨(e1)
= 3 ⋅ 1 + 4 ⋅ 0 + 5 ⋅ 0 = 3.
That’s exactly what we wanted.
You might be inclined to point out that VV∨ at this point, since there’s an obvious
isomorphism eiei∨. You might call it “rotating the column matrix by 90∘”. The issue is that
this isomorphism depends very much on which basis you choose: if I pick a different basis, then
the isomorphism will be intrinsically different.
It is true that V and V∨ are isomorphic for finite-dimensional V , but you should already
know that any two k-vector spaces of the same dimension are isomorphic. In light of this, the
fact that VV∨ is not especially impressive.
11.3V∨⊗ W gives matrices from V to W
Goal of this section:
If V and W are finite-dimensional k-vector spaces then V∨⊗ W represents linearmaps V → W.
Here’s the intuition. If V is three-dimensional and W is five-dimensional, then we can think
of the maps V → W as a 5 × 3 array of numbers. We want to think of these maps as a vector
space: (since one can add or scale matrices). So it had better be a vector space
with dimension 15, but just saying “k⊕15” is not really that satisfying (what is the
basis?).
To do better, we consider the tensor product
which somehow is a product of maps out of V and the target space W. We claim that this is
in fact the space we want: i.e. there is a natural bijection between elements of V∨⊗Wand linear maps from V to W.
First, how do we interpret an element of V∨⊗ W as a map V → W? For concreteness,
suppose V has a basis e1, e2, e3, and W has a basis f1, f2, f3, f4, f5. Consider an element of
V∨⊗ W, say
We want to interpret this element as a function V → W: so given a v ∈ V , we want to
output an element of W. There’s really only one way to do this: feed in v ∈ V into the V∨
guys on the left. That is, take the map
So, there’s a natural way to interpret any element ξ1⊗w1 + + ξm⊗wm∈ V∨⊗W as a
linear map V → W. The claim is that in fact, every linear map V → W has such an
interpretation.
First, for notational convenience,
Definition 11.3.1.Let Hom(V,W) denote the set of linear maps from V to W (which
one can interpret as matrices which send V to W), viewed as a vector space over k. (The
“Hom” stands for homomorphism.)
Question 11.3.2.IdentifyHom(V,k) by name.
We can now write down something that’s more true generally.
Theorem 11.3.3 (V∨⊗ W⇔linear maps V → W) Let V and W be finite-dimensional
vector spaces. We described a map
by sending ξ1⊗ w1 + + ξm⊗ wm to the linear map
Then Ψ is an isomorphism of vector spaces, i.e. every linear map V → W can be uniquely
represented as an element of V∨⊗ W in this way.
The above is perhaps a bit dense, so here is a concrete example.
Example 11.3.4 (Explicit example) Let V = ℝ2 and take a basis e1, e2 of V . Then
define T : V → V by
Then we have
The beauty is that the Ψ definition is basis-free; thus even if we change the basis, although
the above expression will look completely different, the actual element in V∨⊗V doesn’t
change.
Despite this, we’ll indulge ourselves in using coordinates for the proof.
Proof of?? . This looks intimidating, but it’s actually not difficult. We proceed in two
steps:
1.
First, we check that Ψ is surjective; every linear map has at least one representation
in V∨⊗ W. To see this, take any T : V → W. Suppose V has basis e1, e2, e3 and
that T(e1) = w1, T(e2) = w2 and T(e3) = w3. Then the element
works, as it is contrived to agree with T on the basis elements ei.
2.
So it suffices to check now that dimV∨⊗W = dimHom(V,W). Certainly, V∨⊗W
has dimension dimV ⋅dimW. But by viewing Hom(V,W) as dimV ⋅dimW matrices,
we see that it too has dimension dimV ⋅ dimW. □
So there is a natural isomorphism V∨⊗ WHom(V,W). While we did use a basis
liberally in the proof that it works, this doesn’t change the fact that the isomorphism is
“God-given”, depending only on the spirit of V and W itself and not which basis we choose to
express the vector spaces in.
11.4The trace
We are now ready to give the definition of a trace. Recall that a square matrix T can be
thought of as a map T : V → V . According to the above theorem,
so every map V → V can be thought of as an element of V∨⊗V . But we can also define an
evaluation map ev : V∨⊗V → k by “collapsing” each pure tensor: f ⊗vf(v). So this gives
us a composed map
This result is called the trace of a matrix T.
Example 11.4.1 (Example of a trace) Continuing the previous example,
And that is why the trace is the sum of the diagonal entries.
11.5A few harder problems to think about
Problem 11A (Trace is sum of eigenvalues).Let V be an n-dimensional vector space over an
algebraically closed field k. Let T : V → V be a linear map with eigenvalues λ1,
λ2, …, λn (counted with algebraic multiplicity). Show that TrT = λ1 + + λn.
Problem 11B† (Product of traces).Let T : V → V and S : W → W be linear maps
of finite-dimensional vector spaces V and W. Define T ⊗ S : V ⊗ W → V ⊗ W by
v ⊗ wT(v) ⊗ S(w). Prove that
Problem 11C† (Traces kind of commute).Let T : V → W and
S : W → V be linear maps between finite-dimensional vector spaces V and W. Show
that
Problem 11D (Putnam 1988).Let V be an n-dimensional vector space.
Let T : V → V be a linear map and suppose there exists n + 1 eigenvectors, any n of which
are linearly independent. Does it follow that T is a scalar multiple of the identity?
12Determinant
The goal of this chapter is to give the basis-free definition of the determinant: that is, we’re
going to define detT for T : V → V without making reference to the encoding for T. This
will make it obvious the determinant of a matrix does not depend on the choice of
basis, and that several properties are vacuously true (e.g. that the determinant is
multiplicative).
The determinant is only defined for finite-dimensional vector spaces, so if you
want you can restrict your attention to finite-dimensional vector spaces for this
chapter. On the other hand we do not need the ground field to be algebraically
closed.
12.1Wedge product
Prototypical example for this section: Λ2(ℝ2) gives parallelograms.
We’re now going to define something called the wedge product. It will look at first like the
tensor product V ⊗ V , but we’ll have one extra relation.
For simplicity, I’ll first define the wedge product Λ2(V ). But we will later replace 2 with any
n.
Definition 12.1.1.Let V be a k-vector space. The 2-wedge product Λ2(V ) is the abelian
group generated by elements of the form v ∧ w (where v,w ∈ V ), subject to the same
relations
(v1 + v2) ∧ w
= v1∧ w + v2∧ w
v ∧ (w1 + w2)
= v ∧ w1 + v ∧ w2
(c ⋅ v) ∧ w
= v ∧ (c ⋅ w)
plus two additional relations:
As a vector space, its action is given by c ⋅ (v ∧ w) = (c ⋅ v) ∧ w = v ∧ (c ⋅ w).
Exercise 12.1.2.Show that the condition v ∧ w = −(w ∧ v) is actually extraneous: you canderive it from the fact that v ∧ v = 0. (Hint: expand (v + w) ∧ (v + w) = 0.)
This looks almost exactly the same as the definition for a tensor product, with two subtle
differences. The first is that we only have V now, rather than V and W as with the tensor
product1
Secondly, there is a new mysterious relation
What’s that doing there? It seems kind of weird.
I’ll give you a hint.
Example 12.1.3 (Wedge product explicit computation) Let V = ℝ2, and let v = ae1 + be2,
w = ce1 + de2. Now let’s compute v ∧ w in Λ2(V ).
You might know that the area of the parallelogram formed by v and w is ad − bc.
You might recognize it as the determinant of . In fact, you might even know
that the determinant is meant to interpret hypervolumes.
This is absolutely no coincidence. The wedge product is designed to interpret signed
areas. That is, v ∧ w is meant to interpret the area of the parallelogram formed by
v and w. You can see why the condition (cv) ∧ w = v ∧ (cw) would make sense
now. And now of course you know why v ∧ v ought to be zero: it’s an area zero
parallelogram!
The miracle of wedge products is that the only additional condition we need to add to
the tensor product axioms is that v ∧w = −(w ∧v). Then suddenly, the wedge will do all our
work of interpreting volumes for us.
In analog to earlier:
Proposition 12.1.4 (Basis of Λ2(V )) Let V be a vector space with basis e1, …, en. Then
a basis of Λ2(V ) is
where i < j. Hence Λ2(V ) has dimension .
Proof.Surprisingly slippery, and also omitted. (You can derive it from the corresponding
theorem on tensor products.) □
Now I have the courage to define a multi-dimensional wedge product. It’s just the same
thing with more wedges.
Definition 12.1.5.Let V be a vector space and m a positive integer. The space Λm(V )
is generated by wedges of the form
subject to relations
∧ (v1 + v2) ∧…
= (∧ v1∧…) + (∧ v2∧…)
∧ (cv1) ∧ v2∧…
= ∧ v1∧ (cv2) ∧…
∧ v ∧ v ∧…
= 0
∧ v ∧ w ∧…
= −(∧ w ∧ v ∧…)
As a vector space
This definition is pretty wordy, but in English the three conditions say
We should be able to add products like before,
You can put constants onto any of the m components (as is directly pointed out
in the “vector space” action), and
Switching any two adjacent wedges negates the whole wedge.
So this is the natural generalization of Λ2(V ). You can convince yourself that any element of the
form
should still be zero.
Just like e1∧ e2 was a basis earlier, we can find the basis for general m and n.
Proposition 12.1.6 (Basis of the wedge product) Let V be a vector space with basis
e1,…,en. A basis for Λm(V ) consists of the elements
where
Hence Λm(V ) has dimension .
Sketch of proof. We knew earlier that ei1⊗⊗eim was a basis for the tensor product.
Here we have the additional property that (a) if two basis elements re-appear then the
whole thing becomes zero, thus we should assume the i’s are all distinct; and (b) we
can shuffle around elements, and so we arbitrarily decide to put the basis elements in
increasing order. □
12.2The determinant
Prototypical example for this section: (ae1 + be2) ∧ (ce1 + de2) = (ad − bc)(e1∧ e2).
Now we’re ready to define the determinant. Suppose T : V → V is a square matrix. We
claim that the map Λm(V ) → Λm(V ) given on wedges by
and extending linearly to all of Λm(V ) is a linear map. (You can check this yourself if you
like.) We call that map Λm(T).
Example 12.2.1 (Example of Λm(T)) In V = ℝ4 with standard basis e1, e2, e3, e4, let
T(e1) = e2, T(e2) = 2e3, T(e3) = e3 and T(e4) = 2e2 + e3. Then, for example, Λ2(T)
sends
e1∧ e2 + e3∧ e4
T(e1) ∧ T(e2) + T(e3) ∧ T(e4)
= e2∧ 2e3 + e3∧ (2e2 + e3)
= 2(e2∧ e3 + e3∧ e2)
= 0.
Now here’s something interesting. Suppose V has dimension n, and let m = n. Then Λn(V )
has dimension = 1 — it’s a one dimensional space! Hence Λn(V )k.
So Λn(T) can be thought of as a linear map from k to k. But we know that a linear mapfrom k to k is just multiplication by a constant. Hence Λn(T) is multiplication by some
constant.
Definition 12.2.2.Let T : V → V , where V is an n-dimensional vector space. Then
Λn(T) is multiplication by a constant c; we define the determinant of T as c = detT.
Example 12.2.3 (The determinant of a 2 × 2 matrix) Let V = ℝ2 again with basis e1
and e2. Let
In other words, T(e1) = ae1 + be2 and T(e2) = ce1 + de2.
Now let’s consider Λ2(V ). It has a basis e1∧ e2. Now Λ2(T) sends it to
So Λ2(T) : Λ2(V ) → Λ2(V ) is multiplication by detT = ad − bc, because it sent e1∧ e2
to (ad − bc)(e1∧ e2).
And that is the definition of a determinant. Once again, since we defined it in terms of
Λn(T), this definition is totally independent of the choice of basis. In other words, the
determinant can be defined based on T : V → V alone without any reference to
matrices.
Question 12.2.4.Why does Λn(S ∘ T) = Λn(S) ∘ Λn(T)?
In this way, we also get
for free.
More generally if we replace 2 by n, an write out the result of expanding
then you will get the formula
called the Leibniz formula for determinants. American high school students will recognize
it; this is (unfortunately) taught as the definition of the determinant, rather than a corollary
of the better definition using wedge products.
Exercise 12.2.5.Verify that expanding the wedge product yields the Leibniz formula for n = 3.
12.3Characteristic polynomials, and Cayley-Hamilton
Let’s connect with the theory of eigenvalues. Take a map T : V → V , where V is n-dimensional
over an algebraically closed field, and suppose its eigenvalues are λ1, λ2, …, λn (with
repetition). Then the characteristic polynomial is given by
Note that if we’ve written T in Jordan form, that is,
(here each ∗ is either 0 or 1), then we can hack together the definition
The latter definition is what you’ll see in most linear algebra books because it lets you
define the characteristic polynomial without mentioning the word “eigenvalue” (i.e. entirely in
terms of arrays of numbers). I’ll admit it does have the merit that it means that given any
matrix, it’s easy to compute the characteristic polynomial and hence compute the eigenvalues;
but I still think the definition should be done in terms of eigenvalues to begin with. For
instance the determinant definition obscures the following theorem, which is actually a
complete triviality.
Theorem 12.3.1 (Cayley-Hamilton) Let T : V → V be a map of finite-dimensional vector
spaces over an algebraically closed field. Then for any T : V → V , the map pT(T) is the
zero map.
Here, by pT(T) we mean that if
then
is the zero map, where Tk denotes T applied k times. We saw this concept already when we
proved that T had at least one nonzero eigenvector.
Example 12.3.2 (Example of Cayley-Hamilton using determinant definition) Suppose T = . Using the determinant definition of characteristic polynomial, we
find that pT(X) = (X − 1)(X − 4) − (−2)(−3) = X2− 5X − 2. Indeed, you can verify
that
If you define pT without the word eigenvalue, and adopt the evil view that matrices are
arrays of numbers, then this looks like a complete miracle. (Indeed, just look at the terrible
proofs on Wikipedia.)
But if you use the abstract viewpoint of T as a linear map, then the theorem is almost
obvious:
Proof of Cayley-Hamilton. Suppose we write V in Jordan normal form as
where Ji has eigenvalue λi and dimension di. By definition,
By definition, (T −λ1)d1 is the zero map on J1. So pT(T) is zero on J1. Similarly it’s
zero on each of the other Ji’s — end of story. □
Remark 12.3.3 (Tensoring up) — The Cayley-Hamilton theorem holds without the
hypothesis that k is algebraically closed: because for example any real matrix can be
regarded as a matrix with complex coefficients (a trick we’ve mentioned before). I’ll
briefly hint at how you can use tensor products to formalize this idea.
Let’s take the space V = ℝ3, with basis e1, e2, e3. Thus objects in V are of the form
r1e1 +r2e2 +r3e3 where r1, r2, r3 are real numbers. We want to consider essentially the
same vector space, but with complex coefficients zi rather than real coefficients ri.
So here’s what we do: view ℂ as a ℝ-vector space (with basis {1,i}, say) and consider
the complexification
Then you can check that our elements are actually of the form
Here, the tensor product is over ℝ, so we have z⊗rei = (zr)⊗ei for r ∈ ℝ. Then Vℂ can
be thought as a three-dimensional vector space over ℂ, with basis 1⊗ei for i ∈{1,2,3}.
In this way, the tensor product lets us formalize the idea that we “fuse on” complex
coefficients.
If T : V → W is a map, then Tℂ : Vℂ→ Wℂ is just the map z ⊗vz ⊗T(v). You’ll see
this written sometimes as Tℂ = id ⊗ T. One can then apply theorems to Tℂ and try to
deduce the corresponding results on T.
12.4A few harder problems to think about
Problem 12A (Column operations).Show that for any real numbers xij (here 1 ≤i,j ≤ n) we have
Problem 12B (Determinant is product of eigenvalues).Let V be an n-dimensional
vector space over an algebraically closed field k. Let T : V → V be a linear map with
eigenvalues λ1, λ2, …, λn (counted with algebraic multiplicity). Show that detT = λ1…λn.
Problem 12C (Exponential matrix).Let X be an n × n matrix with complex
coefficients. We define the exponential map by
(take it for granted that this converges to some n × n matrix). Prove that
Problem 12D (Extension to ?? ).Let T : V → V be a map of finite-dimensional vector
spaces. Prove that T is an isomorphism if and only if detT≠0.
Problem 12E (Based on Sweden 2010).A herd of 1000 cows of nonzero
weight is given. Prove that we can remove one cow such that the remaining 999 cows cannot
be split into two halves of equal weights.
Problem 12F (Putnam 2015).Define S to be the set of real matrices
such that a, b, c, d form an arithmetic progression in that order. Find all M ∈ S such that for
some integer k > 1, Mk∈ S.
Problem 12G.Let V be a finite-dimensional vector space over k and
T : V → V . Show that
where the trace is taken by viewing Λn(T) : Λn(V ) → Λn(V ).
13Inner product spaces
It will often turn out that our vector spaces which look more like ℝn not only have the notion
of addition, but also a notion of orthogonality and the notion of distance. All this is
achieved by endowing the vector space with a so-called inner form, which you likely
already know as the “dot product” for ℝn. Indeed, in ℝn you already know that
v ⋅ w = 0 if and only if v and w are perpendicular, and
|v|2 = v ⋅ v.
The purpose is to quickly set up this structure in full generality. Some highlights of the chapter:
We’ll see that the high school “dot product” formulation is actually very natural:
it falls out from the two axioms we listed above. If you ever wondered why ∑aibi
behaves as nicely as it does, now you’ll know.
We show how the inner form can be used to make V into a metric space, giving it
more geometric structure.
A few chapters later, we’ll identify VV∨ in a way that wasn’t possible before,
and as a corollary deduce the nice result that symmetric matrices with real entries
always have real eigenvalues.
Throughout this chapter, all vector spaces are over ℂ or ℝ, unless otherwise specified. We’ll
generally prefer working over ℂ instead of ℝ since ℂ is algebraically closed (so, e.g. we have
Jordan forms). Every real matrix can be thought of as a matrix with complex entries
anyways.
13.1The inner product
Prototypical example for this section:Dot product in ℝn.
13.1.iFor real numbers: bilinear forms
First, let’s define the inner form for real spaces. Rather than the notation v ⋅ w it is most
customary to use for general vector spaces.
Definition 13.1.1.Let V be a real vector space. A real inner form1
is a function
which satisfies the following properties:
The form is symmetric: for any v,w ∈ V we have
Of course, one would expect this property from a product.
The form is bilinear, or linear in both arguments, meaning that
and are linear functions for any fixed v. Spelled explicitly this means
that
= c
= + .
and similarly if v was on the left. This is often summarized by the single equation
= c + .
The form is positive definite, meaning ≥ 0 is a nonnegative real number, and
equality takes place only if v = 0V.
Exercise 13.1.2.Show that linearity in the first argument plus symmetry already gives youlinearity in the second argument, so we could edit the above definition by only requiring to be linear.
Example 13.1.3 (ℝn) As we already know, one can define the inner form on ℝn as
follows. Let e1 = (1,0,…,0), e2 = (0,1,…,0), …, en = (0,…,0,1) be the usual basis. Then
we let
It’s easy to see this is bilinear (symmetric and linear in both arguments). To see it is
positive definite, note that if ai = bi then the dot product is a12 + + an2, which is
zero exactly when all ai are zero.
13.1.iiFor complex numbers: sesquilinear forms
The definition for a complex product space is similar, but has one difference: rather than
symmetry we instead have conjugate symmetry meaning =. Thus, while we still
have linearity in the first argument, we actually have a different linearity for the second
argument. To be explicit:
Definition 13.1.4.Let V be a complex vector space. A complex inner product is a
function
which satisfies the following properties:
The form has conjugate symmetry, which means that for any v,w ∈ V we have
The form is sesquilinear (the name means “one-and-a-half linear”). This means
that:
The form is linear in the first argument, so again we have
= +
= c.
Again this is often abbreviated to the single line = c + in the
literature.
However, it is now anti-linearin the second argument: for any complex
number c and vectors x and y we have
= +
=c.
Note the appearance of the complex conjugate c, which is new! Again, we can
abbreviate this to just =c + if we only want to write one
equation.
The form is positive definite, meaning is a nonnegative real number, and equals
zero exactly when v = 0V.
Exercise 13.1.5.Show that anti-linearity follows from conjugate symmetry plus linearity in thefirst argument.
Example 13.1.6 (ℂn) The dot product in ℂn is defined as follows: let e1, e2, …, en be
the standard basis. For complex numbers wi, zi we set
Question 13.1.7.Check that the above is in fact a complex inner form.
13.1.iiiInner product space
It’ll be useful to treat both types of spaces simultaneously:
Definition 13.1.8.An inner product space is either a real vector space equipped
with a real inner form, or a complex vector space equipped with a complex inner form.
A linear map between inner product spaces is a map between the underlying vector
spaces (we do not require any compatibility with the inner form).
Remark 13.1.9 (Why sesquilinear?) — The above example explains one reason why we
want to satisfy conjugate symmetry rather than just symmetry. If we had tried to define
the dot product as ∑wizi, then we would have lost the condition of being positive
definite, because there is no guarantee that = ∑zi2 will even be a real number
at all. On the other hand, with conjugate symmetry we actually enforce =,
i.e. ∈ ℝ for every v.
Let’s make this point a bit more forcefully. Suppose we tried to put a bilinear form
, on a complex vector space V . Let e be any vector with = 1 (a unit vector).
Then we would instead get = − = −1; this is a vector with length −1, which
is not okay! That’s why it is important that, when we have a complex inner product
space, our form is sesquilinear, not bilinear.
Now that we have a dot product, we can talk both about the norm and orthogonality.
13.2Norms
Prototypical example for this section:ℝnbecomes its usual Euclidean space with the vectornorm.
The inner form equips our vector space with a notion of distance, which we call the norm.
Definition 13.2.1.Let V be an inner product space. The norm of v ∈ V is defined by
This definition makes sense because we assumed our form to be positive definite, so
is a nonnegative real number.
Example 13.2.2 (ℝn and ℂn are normed vector spaces) When V = ℝn or V = ℂn with
the standard dot product norm, then the norm of v corresponds to the absolute value
that we are used to.
Our goal now is to prove that
With the metric d(v,w) = , V becomes a metric space.
Question 13.2.3.Verify that d(v,w) = 0 if and only if v = w.
So we just have to establish the triangle inequality. Let’s now prove something we all know
and love, which will be a stepping stone later:
Lemma 13.2.4 (Cauchy-Schwarz) Let V be an inner product space. For any v,w ∈ V we
have
with equality if and only if v and w are linearly dependent.
Proof.The theorem is immediate if = 0. It is also immediate if = 0,
since then one of v or w is the zero vector. So henceforth we assume all these quantities
are nonzero (as we need to divide by them later).
The key to the proof is to think about the equality case: we’ll use the inequality
≥ 0. Deferring the choice of c until later, we compute
0
≤
= −− +
= |c|2− c−c +
= |c|22 + 2− c−c
2Re
≤|c|22 + 2
At this point, a good choice of c is
c
= ⋅
since then
c
= ∈ ℝ
|c|
=
whence the inequality becomes
2
≤ 22
≤.
□
Thus:
Theorem 13.2.5 (Triangle inequality) We always have
with equality if and only if v and w are linearly dependent.
Exercise 13.2.6.Prove this by squaring both sides, and applying Cauchy-Schwarz.
In this way, our vector space now has a topological structure of a metric space.
13.3Orthogonality
Prototypical example for this section:Still ℝn!
Our next goal is to give the geometric notion of “perpendicular”. The definition is easy
enough:
Definition 13.3.1.Two nonzero vectors v and w in an inner product space are
orthogonal if = 0.
As we expect from our geometric intuition in ℝn, this implies independence:
Lemma 13.3.2 (Orthogonal vectors are independent) Any set of pairwise orthogonal
vectors v1, v2, …, vn, with ≠0 for each i, is linearly independent.
Proof.Consider a dependence
for ai in ℝ or ℂ. Then
Hence a1 = 0, since we assumed ≠0. Similarly a2 = = am = 0. □
In light of this, we can now consider a stronger condition on our bases:
Definition 13.3.3.An orthonormal basis of a finite-dimensional inner product space
V is a basis e1, …, en such that = 1 for every i and = 0 for any i≠j.
Example 13.3.4 (ℝn and ℂn have standard bases) In ℝn and ℂn equipped with the
standard dot product, the standard basis e1, …, en is also orthonormal.
This is no loss of generality:
Theorem 13.3.5 (Gram-Schmidt) Let V be a finite-dimensional inner product space.
Then it has an orthonormal basis.
Sketch of Proof. One constructs the orthonormal basis explicitly from any basis e1, …, en of
V . Define proju(v) = u. Then recursively define
u1
= e1
u2
= e2− proju1(e2)
u3
= e3− proju1(e3) − proju2(e3)
un
= en− proju1(en) −− projun−1(en).
One can show the ui are pairwise orthogonal and not zero. □
Thus, we can generally assume our bases are orthonormal.
Worth remarking:
Example 13.3.6 (The dot product is the “only” inner form) Let V be a finite-dimensional inner product space, and consider any orthonormal basis
e1,…,en. Then we have that
owing to the fact that the {ei} are orthonormal.
And now you know why the dot product expression is so ubiquitous.
13.4Hilbert spaces
In algebra we are usually scared of infinity, and so when we defined a basis of a vanilla vector
space many chapters ago, we only allowed finite linear combinations. However, if we have an
inner product space, then it is a metric space and we can sometimes actually talk about
convergence.
Here is how it goes:
Definition 13.4.1.A Hilbert space is a inner product space V , such that the
corresponding metric space is complete.
In that case, it will now often make sense to take infinite linear combinations, because we
can look at the sequence of partial sums and let it converge. Here is how we might do it. Let’s
suppose we have e1, e2, …an infinite sequence of vectors with norm 1 and which are pairwise
orthogonal. Suppose c1, c2, …, is a sequence of real or complex numbers. Then consider the
sequence
v1
= c1e1
v2
= c1e1 + c2e2
v3
= c1e1 + c2e2 + c3e3
Proposition 13.4.2 (Convergence criteria in a Hilbert space) The sequence (vi) defined
above converges if and only if ∑2< ∞.
Proof.This will make more sense if you read ?? , so you could skip this proof if you
haven’t read the chapter. The sequence vi converges if and only if it is Cauchy, meaning
that when i < j,
tends to zero as i and j get large. This is equivalent to the sequence sn = |c1|2 ++
|cn|2 being Cauchy.
Since ℝ is complete, sn is Cauchy if and only if it converges. Since sn consists of
nonnegative real numbers, converges holds if and only if sn is bounded, or equivalently
if ∑2< ∞. □
Thus, when we have a Hilbert space, we change our definition slightly:
Definition 13.4.3.An orthonormal basis for a Hilbert space V is a (possibly infinite)
sequence e1, e2, …, of vectors such that
= 1 for all i,
= 0 for i≠j, i.e. the vectors are pairwise orthogonal
every element of V can be expressed uniquely as an infinite linear combination
where ∑i2< ∞, as described above.
That’s the official definition, anyways. (Note that if dimV < ∞, this agrees with our usual
definition, since then there are only finitely many ei.) But for our purposes you can mostly not
worry about it and instead think:
A Hilbert space is an inner product space whose basis requires infinite linearcombinations, not just finite ones.
The technical condition ∑2< ∞ is exactly the one which ensures the infinite sum
makes sense.
13.5A few harder problems to think about
Problem 13A (Pythagorean theorem).Show that if = 0 in an inner product
space, then 2 + 2 = 2.
Problem 13B⋆ (Finite-dimensional Hilbert).Show that a finite-dimensional inner
product space is a Hilbert space.
Problem 13C (Taiwan IMO camp).In a town there are n people and k
clubs. Each club has an odd number of members, and any two clubs have an even number of
common members. Prove that k ≤ n.
Problem 13D⋆ (Inner product structure of tensors). Let V and W be finite-dimensional
inner product spaces over k, where k is either ℝ or ℂ.
(a)
Find a canonical way to make V ⊗kW into an inner product space too.
(b)
Let e1, …, en be an orthonormal basis of V and f1, …, fm be an orthonormal basis
of W. What’s an orthonormal basis of V ⊗ W?
Problem 13E (Putnam 2014).Let n be a positive integer. What is the
largest k for which there exist n×n matrices M1,…,Mk and N1,…,Nk with real entries such
that for all i and j, the matrix product MiNj has a zero entry somewhere on its diagonal if
and only if i≠j?
Problem 13F (Sequence space).Consider the space ℓ2 of infinite sequences of real
numbers a = (a1,a2,…) satisfying ∑iai2< ∞. We equip it with the dot product
Is this a Hilbert space? If so, identify a Hilbert basis.
Problem 13G (Kuratowski embedding).A Banach space is a normed vector space
V , such that the corresponding metric space is complete. (So a Hilbert space is a special
case of a Banach space.)
Let (M,d) be any metric space. Prove that there exists a Banach space X and an
injective function f : MX such that d(x,y) = for any x and y.
14Bonus: Fourier analysis
Now that we’ve worked hard to define abstract inner product spaces, I want to
give an (optional) application: how to set up Fourier analysis correctly, using this
language.
For fun, I also prove a form of Arrow’s Impossibility Theorem using binary Fourier
analysis.
In what follows, we let 𝕋 = ℝ∕ℤ denote the “circle group”, thought of as the additive group
of “real numbers modulo 1”. There is a canonical map e: 𝕋 → ℂ sending 𝕋 to the complex
unit circle, given by
14.1Synopsis
Suppose we have a domain Z and are interested in functions f : Z → ℂ. Naturally, the set of
such functions form a complex vector space. We like to equip the set of such functions with an
positive definite inner product.
The idea of Fourier analysis is to then select an orthonormal basis for this set of functions,
say (eξ)ξ, which we call the characters; the indexing ξ are called frequencies. In that case,
since we have a basis, every function f : Z → ℂ becomes a sum
where f(ξ) are complex coefficients of the basis; appropriately we call f the Fouriercoefficients. The variable x ∈ Z is referred to as the physical variable. This is generally good
because the characters are deliberately chosen to be nice “symmetric” functions, like sine or
cosine waves or other periodic functions. Thus we decompose an arbitrarily complicated
function into a sum of nice ones.
14.2A reminder on Hilbert spaces
For convenience, we record a few facts about orthonormal bases.
Proposition 14.2.1 (Facts about orthonormal bases) Let V be a complex Hilbert space with
inner form and suppose x = ∑ξaξeξ and y = ∑ξbξeξ where eξ are an orthonormal
basis. Then
= ∑ξ|aξ|2
aξ
=
= ∑ξaξbξ.
Exercise 14.2.2.Prove all of these. (You don’t need any of the preceding section, it’s only thereto motivate the notation with lots of scary ξ’s.)
In what follows, most of the examples will be of finite-dimensional inner product spaces
(which are thus Hilbert spaces), but the example of “square-integrable functions” will
actually be an infinite dimensional example. Fortunately, as I alluded to earlier,
this is no cause for alarm and you can mostly close your eyes and not worry about
infinity.
14.3Common examples
14.3.iBinary Fourier analysis on {±1}n
Let Z = {±1}n for some positive integer n, so we are considering functions f(x1,…,xn)
accepting binary values. Then the functions Z → ℂ form a 2n-dimensional vector space ℂZ,
and we endow it with the inner form
In particular,
is the average of the squares; this establishes also that is positive definite.
In that case, the multilinear polynomials form a basis of ℂZ, that is the polynomials
Exercise 14.3.1.Show that they’re actually orthonormal under . This proves they forma basis, since there are 2nof them.
Thus our frequency set is actually the subsets S ⊆{1,…,n}. Thus, we have a
decomposition
Example 14.3.2 (An example of binary Fourier analysis) Let n = 2. Then binary
functions {±1}2→ ℂ have a basis given by the four polynomials
For example, consider the function f which is 1 at (1,1) and 0 elsewhere. Then we can
put
So the Fourier coefficients are f(S) = for each of the four S’s.
This notion is useful in particular for binary functions f : {±1}n→{±1}; for these
functions (and products thereof), we always have = 1.
It is worth noting that the frequency ∅ plays a special role:
Exercise 14.3.3.Show that
14.3.iiFourier analysis on finite groups Z
This time, suppose we have a finite abelian group Z, and consider functions Z → ℂ; this is a
|Z|-dimensional vector space. The inner product is the same as before:
To proceed, we’ll need to be able to multiply two elements of Z. This is a bit of a nuisance
since it actually won’t really matter what map I pick, so I’ll move briskly; feel free to skip
most or all of the remaining paragraph.
Definition 14.3.4.We select a symmetric non-degenerate bilinear form
satisfying the following properties:
ξ ⋅ (x1 + x2) = ξ ⋅ x1 + ξ ⋅ x2 and (ξ1 + ξ2) ⋅ x = ξ1⋅ x + ξ2⋅ x (this is the word
“bilinear”)
⋅ is symmetric,
For any ξ≠0, there is an x with ξ ⋅ x≠0 (this is the word “nondegenerate”).
Example 14.3.5 (The form on ℤ∕nℤ) If Z = ℤ∕nℤ then ξ ⋅ x = (ξx)∕n satisfies the
above.
In general, it turns out finite abelian groups decompose as the sum of cyclic groups (see
?? ), which makes it relatively easy to find such a ⋅; but as I said the choice won’t matter, so
let’s move on.
Now for the fun part: defining the characters.
Proposition 14.3.6 (eξare orthonormal) For each ξ ∈ Z we define the character
The |Z| characters form an orthonormal basis of the space of functions Z → ℂ.
Proof.I recommend skipping this one, but it is:
= ∑x∈Ze(ξ ⋅ x)e(ξ′⋅ x)
= ∑x∈Ze(ξ ⋅ x)e(−ξ′⋅ x)
= ∑x∈Ze.
□
In this way, the set of frequencies is also Z, but the ξ ∈ Z play very different roles from the
“physical” x ∈ Z. Here is an example which might be enlightening.
Example 14.3.7 (Cube roots of unity filter) Suppose Z = ℤ∕3ℤ, with the inner form
given by ξ ⋅ x = (ξx)∕3. Let ω = exp(πi) be a primitive cube root of unity. Note that
Then given f : Z → ℂ with f(0) = a, f(1) = b, f(2) = c, we obtain
In this way we derive that the transforms are
f(0)
=
f(1)
=
f(2)
= .
Exercise 14.3.8.Show that in analogy tof(∅) for binary Fourier analysis, we now have
Olympiad contestants may recognize the previous example as a “roots of unity filter”, which
is exactly the point. For concreteness, suppose one wants to compute
In that case, we can consider the function
such that w(0) = 1 but w(1) = w(2) = 0. By abuse of notation we will also think of w as a
function w : ℤ ↠ ℤ∕3 → ℂ. Then the sum in question is
∑nw(n)
= ∑n∑k=0,1,2w(k)ωkn
= ∑k=0,1,2w(k)∑nωkn
= ∑k=0,1,2w(k)(1 + ωk)n.
In our situation, we have w(0) = w(1) = w(2) = , and we have evaluated the desired sum.
More generally, we can take any periodic weight w and use Fourier analysis in order to
interchange the order of summation.
Example 14.3.9 (Binary Fourier analysis) Suppose Z = {±1}n, viewed as an abelian
group under pointwise multiplication hence isomorphic to (ℤ∕2ℤ)⊕n. Assume we pick
the dot product defined by
where ξ = (ξ1,…,ξn) and x = (x1,…,xn).
We claim this coincides with the first example we gave. Indeed, let S ⊆{1,…,n} and let
ξ ∈{±1}n which is −1 at positions in S, and +1 at positions not in S. Then the character
χS from the previous example coincides with the character eξ in the new notation. In
particular, f(S) = f(ξ).
Thus Fourier analysis on a finite group Z subsumes binary Fourier analysis.
14.3.iiiFourier series for functions L2([−π,π])
This is the most famous one, and hence the one you’ve heard of.
Definition 14.3.10.The space L2([−π,π]) consists of all functions f : [−π,π] → ℂ
such that the integral ∫[−π,π]2dx exists and is finite, modulo the relation that a
function which is zero “almost everywhere” is considered to equal zero.1
It is made into an inner product space according to
It turns out (we won’t prove) that this is an (infinite-dimensional) Hilbert space!
Now, the beauty of Fourier analysis is that this space has a great basis:
Theorem 14.3.11 (The classical Fourier basis) For each integer n, define
Then en form an orthonormal basis of the Hilbert space L2([−π,π]).
Thus this time the frequency set ℤ is infinite, and we have
for coefficients f(n) with ∑n2< ∞. Since the frequency set is indexed by ℤ, we call
this a Fourier series to reflect the fact that the index is n ∈ ℤ.
Exercise 14.3.12.Show once again
14.4Summary, and another teaser
We summarize our various flavors of Fourier analysis in the following table.
I snuck in a fourth row with Z = ℤ∕nℤ, but it’s a special case of the second row, so no cause
for alarm.
Alluding to the future, I want to hint at how ?? starts. Each one of these is
really a statement about how functions from G → ℂ can be expressed in terms of
functions G→ ℂ, for some “dual” G. In that sense, we could rewrite the above table
as:
It will turn out that in general we can say something about many different domains G, once
we know what it means to integrate a measure. This is the so-called Pontryagin duality; and it
is discussed as a follow-up bonus in ?? .
14.5Parseval and friends
Here is a fun section in which you get to learn a lot of big names quickly. Basically, we can
take each of the three results from Proposition 14.2.1, translate it into the context of our
Fourier analysis (for which we have an orthonormal basis of the Hilbert space), and get a
big-name result.
Corollary 14.5.1 (Parseval theorem) Let f : Z → ℂ, where Z is a finite abelian group.
Then
Similarly, if f : [−π,π] → ℂ is square-integrable then its Fourier series satisfies
Proof.Recall that is equal to the square sum of the coefficients. □
Corollary 14.5.2 (Fourier inversion formula) Let f : Z → ℂ, where Z is a finite abelian
group. Then
Similarly, if f : [−π,π] → ℂ is square-integrable then its Fourier series is given by
Proof.Recall that in an orthonormal basis (eξ)ξ, the coefficient of eξ in f is . □
Question 14.5.3.What happens when ξ = 0 above?
Corollary 14.5.4 (Plancherel theorem) Let f : Z → ℂ, where Z is a finite abelian group.
Then
Similarly, if f : [−π,π] → ℂ is square-integrable then
Question 14.5.5.Prove this one in one line (like before).
14.6Application: Basel problem
One cute application about Fourier analysis on L2([−π,π]) is that you can get some otherwise
hard-to-compute sums, as long as you are willing to use a little calculus.
Here is the classical one:
Theorem 14.6.1 (Basel problem) We have
The proof is to consider the identity function f(x) = x, which is certainly square-integrable.
Then by Parseval, we have
A calculus computation gives
On the other hand, we will now compute all Fourier coefficients. We have already
that
For n≠0, we have by definition (or “Fourier inversion formula”, if you want to use big words)
the formula
f(n)
=
= ∫[−π,π]x ⋅exp(inx)dx
= ∫[−π,π]xexp(−inx) dx.
The anti-derivative is equal to exp(−inx)(1 + inx), which thus with some more
calculation gives that
So
implying the result.
14.7Application: Arrow’s Impossibility Theorem
As an application of binary Fourier analysis, we now prove a form of Arrow’s theorem.
Consider n voters voting among 3 candidates A, B, C. Each voter specifies a tuple
vi = (xi,yi,zi) ∈{±1}3 as follows:
xi = 1 if person i ranks A ahead of B, and xi = −1 otherwise.
yi = 1 if person i ranks B ahead of C, and yi = −1 otherwise.
zi = 1 if person i ranks C ahead of A, and zi = −1 otherwise.
Tacitly, we only consider 3! = 6 possibilities for vi: we forbid “paradoxical” votes of the form
xi = yi = zi by assuming that people’s votes are consistent (meaning the preferences are
transitive).
For brevity, let x∙ = (x1,…,xn) and define y∙ and z∙ similarly. Then, we can consider a
voting mechanism
f : {±1}n
→{±1}
g: {±1}n
→{±1}
h: {±1}n
→{±1}
such that
f(x∙) is the global preference of A vs. B,
g(y∙) is the global preference of B vs. C,
and h(z∙) is the global preference of C vs. A.
We’d like to avoid situations where the global preference (f(x∙),g(y∙),h(z∙)) is itself
paradoxical.
Let 𝔼f denote the average value of f across all 2n inputs. Define 𝔼g and 𝔼h similarly. We’ll
add an assumption that 𝔼f = 𝔼g = 𝔼h = 0, which provides symmetry (and e.g. excludes the
possibility that f, g, h are constant functions which ignore voter input). With that we will
prove the following result:
Theorem 14.7.1 (Arrow Impossibility Theorem) Assume that (f,g,h) always avoids
paradoxical outcomes, and assume 𝔼f = 𝔼g = 𝔼h = 0. Then (f,g,h) is either a
dictatorship or anti-dictatorship: there exists a “dictator” k such that
where all three signs coincide.
Unlike the usual Arrow theorem, we do not assume that f(+1,…,+1) = +1 (hence
possibility of anti-dictatorship).
Proof.Suppose the voters each randomly select one of the 3! = 6 possible consistent
votes. In ?? it is shown that the exact probability of a paradoxical outcome for any
functions f, g, h is given exactly by
Assume that this probability (of a paradoxical outcome) equals 0. Then, we derive
But now we can just use weak inequalities. We have f(∅) = 𝔼f = 0 and similarly for g
and h, so we restrict attention to |S|≥ 1. We then combine the famous inequality
|ab + bc + ca|≤ a2 + b2 + c2 (which is true across all real numbers) to deduce that
1
= ∑S⊆{1,…,n}−
≤∑S⊆{1,…,n}
≤∑S⊆{1,…,n}1
= (1 + 1 + 1) = 1.
with the last step by Parseval. So all inequalities must be sharp, and in particular f, g, h
are supported on one-element sets, i.e. they are linear in inputs. As f, g, h are ±1 valued,
each f, g, h is itself either a dictator or anti-dictator function. Since (f,g,h) is always
consistent, this implies the final result. □
14.8A few harder problems to think about
Problem 14A (For calculus fans).Prove that
Problem 14B. Let f,g,h: {±1}n→{±1} be any three functions.
For each i, we randomly select (xi,yi,zi) ∈{±1}3 subject to the constraint that not all
are equal (hence, choosing among 23− 2 = 6 possibilities). Prove that the probability
that
is given by the formula
15Duals, adjoint, and transposes
This chapter is dedicated to the basis-free interpretation of the transpose and conjugate
transpose of a matrix.
Poster corollary: we will see that symmetric matrices with real coefficients are diagonalizable
and have real eigenvalues.
15.1Dual of a map
Prototypical example for this section:The example below.
We go ahead and now define a notion that will grow up to be the transpose of a
matrix.
Definition 15.1.1.Let V and W be vector spaces. Suppose T : V → W is a linear map. Then
we actually get a map
T∨: W∨
→ V∨
f
f ∘ T.
This map is called the dual map.
Example 15.1.2 (Example of a dual map) Work over ℝ. Let’s consider V with basis e1, e2, e3
and W with basis f1, f2. Suppose that
T(e1)
= f1 + 2f2
T(e2)
= 3f1 + 4f2
T(e3)
= 5f1 + 6f2.
Now consider V∨ with its dual basis e1∨, e2∨, e3∨ and W∨ with its dual basis f1∨, f2∨. Let’s
compute T∨(f1∨) = f1∨∘ T: it is given by
f1∨
= f1∨
= a + 3b + 5c.
So accordingly we can write
Similarly,
This determines T∨ completely.
If we write the matrices for T and T∨ in terms of our basis, we now see that
So in our selected basis, we find that the matrices are transposes: mirror images of each
other over the diagonal.
Of course, this should work in general.
Theorem 15.1.3 (Transpose interpretation of T∨) Let V and W be finite-dimensional
k-vector spaces. Then, for any T : V → W, the following two matrices are transposes:
The matrix for T : V → W expressed in the basis (ei), (fj).
The matrix for T∨: W∨→ V∨ expressed in the basis (fj∨), (ei∨).
Proof.The (i,j)th entry of the matrix T corresponds to the coefficient of fj in T(ei),
which corresponds to the coefficient of ei∨ in fj∨∘ T. □
The nice part of this is that the definition of T∨ is basis-free. So it means that if we start
with any linear map T, and then pick whichever basis we feel like, then T and T∨ will still be
transposes.
15.2Identifying with the dual space
For the rest of this chapter, though, we’ll now bring inner products into the picture.
Earlier I complained that there was no natural isomorphism VV∨. But in fact, given an
inner form we can actually make such an identification: that is we can naturally associate
every linear map ξ: V → k with a vector v ∈ V .
To see how we might do this, suppose V = ℝ3 for now with an orthonormal basis
e1, e2, e3. How might we use the inner product to represent a map from V → ℝ?
For example, take ξ ∈ V∨ by ξ(e1) = 3, ξ(e2) = 4 and ξ(e3) = 5. Actually, I claim
that
for every v.
Question 15.2.1.Check this.
And this works beautifully in the real case.
Theorem 15.2.2 (VV∨for real inner form) Let V be a finite-dimensional real inner
product space and V∨ its dual. Then the map V → V∨ by
is an isomorphism of real vector spaces.
Proof.It suffices to show that the map is injective and surjective.
Injective: suppose = for every vector v ∈ V . This means
= 0 for every vector v ∈ V . This can only happen if v1− v2 = 0; for
example, take v = v1− v2 and use positive definiteness.
Surjective: take an orthonormal basis e1, …en and let e1∨, …, en∨ be the dual basis
on V∨. Then e1 maps to e1∨, et cetera. □
Actually, since we already know dimV = dimV∨ we only had to prove one of the above. As
a matter of personal taste, I find the proof of injectivity more elegant, and the proof of
surjectivity more enlightening, so I included both. Thus
If a real inner product space V is given an inner form, then V and V∨arecanonically isomorphic.
Unfortunately, things go awry if V is complex. Here is the results:
Theorem 15.2.3 (V versus V∨for complex inner forms) Let V be a finite-dimensional
complex inner product space and V∨ its dual. Then the map V → V∨ by
is a bijection of sets.
Wait, what? Well, the proof above shows that it is both injective and surjective, but why is
it not an isomorphism? The answer is that it is not a linear map: since the form is sesquilinear
we have for example
which has introduced a minus sign! In fact, it is an anti-linear map, in the sense we defined
before.
Eager readers might try to fix this by defining the isomorphism v instead. However,
this also fails, because the right-hand side is not even an element of V∨: it is an “anti-linear”,
not linear.
And so we are stuck. Fortunately, we will only need the “bijection” result for what
follows, so we can continue on anyways. (If you want to fix this, ?? gives a way to do
so.)
15.3The adjoint (conjugate transpose)
We will see that, as a result of the flipping above, the conjugate transpose is actually the
better concept for inner product spaces: since it can be defined using only the inner product
without making mention to dual spaces at all.
Definition 15.3.1.Let V and W be finite-dimensional inner product spaces, and let
T : V → W. The adjoint (or conjugate transpose) of T, denoted T†: W → V , is
defined as follows: for every vector w ∈ W, we let T†(w) ∈ V be the unique vector with
for every v ∈ V .
Some immediate remarks about this definition:
Our T† is well-defined, because W is some function in V∨, and hence by
the bijection earlier it should be uniquely of the form for some v ∈ V .
This map T† is indeed a linear map (why?).
The niceness of this definition is that it doesn’t make reference to any basis or even
V∨, so it is the “right” definition for a inner product space.
By symmetry, of course, we also have = .
Example 15.3.2 (Example of an adjoint map) We’ll work over ℂ, so the conjugates are more
visible. Let’s consider V with orthonormal basis e1, e2, e3 and W with orthonormal basis f1,
f2. We put
T(e1)
= if1 + 2f2
T(e2)
= 3f1 + 4f2
T(e3)
= 5f1 + 6if2.
We compute T†(f1). It is the unique vector x ∈ V such that
for any v ∈ V . If we expand v = ae1 + be2 + ce3 the above equality becomes
V
= W
= ia + 3b + 5c.
However, since x is in the second argument, this means we actually want to take
so that the sesquilinearity will conjugate the i.
The pattern continues, though we remind the reader that we need the basis to be
orthonormal to proceed.
Theorem 15.3.3 (Adjoints are conjugate transposes) Fix an orthonormal basis of a
finite-dimensional inner product space V . Let T : V → V be a linear map. If we write
T as a matrix in this basis, then the matrix T† (in the same basis) is the conjugatetranspose of the matrix of T; that is, the (i,j)th entry of T∨ is the complex conjugate
of the (j,i)th entry of T.
Proof.One-line version: take v and w to be basis elements, and this falls right out.
Full proof: let
in this basis e1, …, en. Then, letting w = ei and v = ej we deduce that
for any i, which is enough to deduce the result. □
15.4Eigenvalues of normal maps
We now come to the advertised theorem. Restrict to the situation where T : V → V . You see,
the world would be a very beautiful place if it turned out that we could pick a basis of
eigenvectors that was also orthonormal. This is of course far too much to hope for; even
without the orthonormal condition, we saw that Jordan form could still have 1’s off the
diagonal.
However, it turns out that there is a complete characterization of exactly when our
overzealous dream is true.
Definition 15.4.1.We say a linear map T (from a finite-dimensional inner product
space to itself) is normal if TT† = T†T.
We say a complex T is self-adjoint or Hermitian if T = T†; i.e. as a matrix in any
orthonormal basis, T is its own conjugate transpose. For real T we say “self-adjoint”,
“Hermitian” or symmetric.
Theorem 15.4.2 (Normal⇔diagonalizable with orthonormal basis) Let V be
a finite-dimensional complex inner product space. A linear map T : V → V is normal if
and only if one can pick an orthonormal basis of eigenvectors.
Exercise 15.4.3.Show that if there exists such an orthonormal basis then T:V → V is normal,by writing T as a diagonal matrix in that basis.
Proof.This is long, and maybe should be omitted on a first reading. If T has an
orthonormal basis of eigenvectors, this result is immediate.
Now assume T is normal. We first prove T is diagonalizable; this is the hard part.
Claim 15.4.4.If T is normal, then kerT = kerTr = kerT† for r ≥ 1. (Here Tr is T
applied r times.)
Proof of Claim. Let S = T†∘T, which is self-adjoint. We first note that S is Hermitian
and kerS = kerT. To see it’s Hermitian, note = = . Taking
v = w also implies kerS ⊆ kerT (and hence equality since obviously kerT ⊆ kerS).
First, since we have = , an induction shows that
kerS = kerSr for r ≥ 1. Now, since T is normal, we have Sr = (T†)r∘ Tr, and thus we
have the inclusion
where the last equality follows from the first claim. Thus in fact kerT = kerTr.
Finally, to show equality with kerT† we
=
=
= .
■
Now consider the given T, and any λ.
Question 15.4.5.Show that (T − λid)†= T†−λid. Thus if T is normal, so is T − λid.
In particular, for any eigenvalue λ of T, we find that ker(T − λid) = ker(T − λid)r. This
implies that all the Jordan blocks of T have size 1; i.e. that T is in fact diagonalizable. Finally,
we conclude that the eigenvectors of T and T† match, and the eigenvalues are complex
conjugates.
So, diagonalize T. We just need to show that if v and w are eigenvectors of T with distinct
eigenvalues, then they are orthogonal. (We can use Gram-Schmidt on any eigenvalue that
appears multiple times.) To do this, suppose T(v) = λv and T(w) = μw (thus T†(w) =μw).
Then
Since λ≠μ, we conclude = 0. □
This means that not only can we write
but moreover that the basis associated with this matrix happens to be orthonormal
vectors.
As a corollary:
Theorem 15.4.6 (Hermitian matrices have real eigenvalues) A Hermitian matrix T is
diagonalizable, and all its eigenvalues are real.
Proof.Obviously Hermitian normal, so write it in the orthonormal basis of
eigenvectors. To see that the eigenvalues are real, note that T = T† means λi =λi for
every i. □
15.5A few harder problems to think about
Problem 15A⋆ (Double dual). Let V be a finite-dimensional vector space.
Prove that
V
→ (V∨)∨
v
gives an isomorphism. (This is significant because the isomorphism is canonical, and
in particular does not depend on the choice of basis. So this is more impressive.)
Problem 15B (Fundamental theorem of linear algebra).Let T : V → W be a map of
finite-dimensional k-vector spaces. Prove that
Problem 15C† (Row rank is column rank).A m×n matrix M of real numbers is given. The
column rank of M is the dimension of the span in ℝm of its n column vectors. The row rank of
M is the dimension of the span in ℝn of its m row vectors. Prove that the row rank and
column rank are equal.
Problem 15D (The complex conjugate spaces). Let V = (V,+,⋅) be a complex vector
space. Define the complex conjugate vector space, denoted V = (V,+,∗) by
changing just the multiplication:
Show that for any sesquilinear form on V , if V is finite-dimensional, then
V
→ V∨
v
is an isomorphism of complex vector spaces.
Problem 15E (T† vs T∨).Let V and W be real inner product spaces and let T : V → W be
an inner product. Show that the following diagram commutes:
Here the isomorphisms are v. Thus, for real inner product spaces, T† is just T∨ with the
duals eliminated (by ?? ).
Problem 15F (Polynomial criteria for normality).Let V be a complex inner product
space and let T : V → V be a linear map. Show that T is normal if and only if there is a
polynomial1p ∈ ℂ[t] such that
Part V More on Groups
16Group actions overkill AIME problems
Consider this problem from the 1996 AIME:
(AIME 1996) Two of the squares of a 7 × 7 checkerboard are painted yellow,
and the rest are painted green. Two color schemes are equivalent if one can
be obtained from the other by applying a rotation in the plane of the board.
How many inequivalent color schemes are possible?
What’s happening here? Let X be the set of the possible colorings of the board. What’s
the natural interpretation of “rotation”? Answer: the group ℤ∕4ℤ = somehow
“acts” on this set X by sending one state x ∈ X to another state r ⋅x, which is just x rotated
by 90∘. Intuitively we’re just saying that two configurations are the same if they can be
reached from one another by this “action”.
We can make all of this precise using the idea of a group action.
16.1Definition of a group action
Prototypical example for this section:The AIME problem.
Definition 16.1.1.Let X be a set and G a group. A group action is a binary operation
⋅ : G × X → X which lets a g ∈ G send an x ∈ X to g ⋅ x. It satisfies the axioms
(g1g2) ⋅ x = g1⋅ (g2⋅ x) for any g1,g2∈ G for all x ∈ X.
1G⋅ x = x for any x ∈ X.
Example 16.1.2 (Examples of group actions) Let G = (G,⋆) be a group.
(a)
The group ℤ∕4ℤ can act on the set of ways to color a 7 × 7 board either yellow or
green.
(b)
The group ℤ∕4ℤ = acts on the xy-plane ℝ2 as follows: r⋅(x,y) = (y,−x).
In other words, it’s a rotation by 90∘.
(c)
The dihedral group D2n acts on the set of ways to color the vertices of an n-gon.
(d)
The group Sn acts on X = by applying the permutation σ: σ⋅x : = σ(x).
(e)
The group G can act on itself (i.e. X = G) by left multiplication: put g ⋅g′: = g ⋆g′.
16.2Stabilizers and orbits
Prototypical example for this section:Again the AIME problem.
Given a group action G on X, we can define an equivalence relation ∼ on X as follows:
x ∼ y if x = g ⋅ y for some g ∈ G. For example, in the AIME problem, ∼ means “one can be
obtained from the other by a rotation”.
Question 16.2.1.Why is this an equivalence relation?
In that case, the AIME problem wants the number of equivalence classes under ∼. So
let’s give these equivalence classes a name: orbits. We usually denote orbits by
𝒪.
As usual, orbits carve out X into equivalence classes.
It turns out that a very closely related concept is:
Definition 16.2.2.The stabilizer of a point x ∈ X, denoted StabG(x), is the set of
g ∈ G which fix x; in other words
Example 16.2.3 Consider the AIME problem again, with X the possible set of states
(again G = ℤ∕4ℤ). Let x be the configuration where two opposite corners are colored
yellow. Evidently 1G fixes x, but so does the 180∘ rotation r2. But r and r3 do not
preserve x, so StabG(x) = {1,r2}ℤ∕2ℤ.
Question 16.2.4.Why isStabG(x) a subgroup of G?
Once we realize the stabilizer is a group, this leads us to what I privately call the
“fundamental theorem of how big an orbit is”.
Theorem 16.2.5 (Orbit-stabilizer theorem) Let 𝒪 be an orbit, and pick any x ∈ 𝒪. Let
S = StabG(x) be a subgroup of G. There is a natural bijection between 𝒪 and left
cosets. In particular,
In particular, the stabilizers of each x ∈𝒪 have the same size.
Proof.The point is that every coset gS just specifies an element of 𝒪, namely g ⋅x. The
fact that S is a stabilizer implies that it is irrelevant which representative we pick.
Since the cosets partition G, each of size , we obtain the second result. □
16.3Burnside’s lemma
Now for the crux of this chapter: a way to count the number of orbits.
Theorem 16.3.1 (Burnside’s lemma) Let G act on a set X. The number of orbits of the
action is equal to
where FixPtg is the set of points x ∈ X such that g ⋅ x = x.
The proof is deferred as a bonus problem, since it has a very olympiad-flavored
solution. As usual, this lemma was not actually proven by Burnside; Cauchy got there
first, and thus it is sometimes called the lemma that is not Burnside’s. Example
application:
Example 16.3.2 (AIME 1996) Two of the squares of a 7 ×7 checkerboard are painted yellow,and the rest are painted green. Two color schemes are equivalent if one can be obtained from the otherby applying a rotation in the plane of the board. How many inequivalent color schemes are possible?
We know that G = ℤ∕4ℤ acts on the set X of possible coloring schemes. Now we can
compute FixPtg explicitly for each g ∈ ℤ∕4ℤ.
If g = 1G, then every coloring is fixed, for a count of = 1176.
If g = r2 there are exactly 24 coloring schemes fixed by g: this occurs when the two
squares are reflections across the center, which means they are preserved under a
180∘ rotation.
If g = r or g = r3, then there are no fixed coloring schemes.
As = 4, the average is
Exercise 16.3.3 (MathCounts Chapter Target Round).A circular spinner has seven sections ofequal size, each of which is colored either red or blue. Two colorings are considered the same ifone can be rotated to yield the other. In how many ways can the spinner be colored? (Answer:20)
Consult [?] for some more examples of “hands-on” applications.
16.4Conjugation of elements
Prototypical example for this section:In Sn, conjugacy classes are “cycle types”.
A particularly common type of action is the so-called conjugation. We let G act on itself
as follows:
You might think this definition is a little artificial. Who cares about the element ghg−1? Let
me try to convince you this definition is not so unnatural.
Example 16.4.1 (Conjugacy in Sn) Let G = S5, and fix a π ∈ S5. Here’s the question:
is πσπ−1 related to σ? To illustrate this, I’ll write out a completely random example of
a permutation σ ∈ S5.
Thus our fixed π doesn’t really change the structure of σ at all: it just “renames” each
of the elements 1, 2, 3, 4, 5 to π(1), π(2), π(3), π(4), π(5).
But wait, you say. That’s just a very particular type of group behaving nicely under
conjugation. Why does this mean anything more generally? All I have to say is: remember
Cayley’s theorem! (This was ?? .)
In any case, we may now define:
Definition 16.4.2.The conjugacy classes of a group G are the orbits of G under the
conjugacy action.
Let’s see what the conjugacy classes of Sn are, for example.
Example 16.4.3 (Conjugacy classes of Sn correspond to cycle types) Intuitively,
the discussion above says that two elements of Sn should be conjugate if they have the
same “shape”, regardless of what the elements are named. The right way to make the
notion of “shape” rigorous is cycle notation. For example, consider the permutation
in cycle notation, meaning 1351 and 242. It is conjugate to the
permutation
or any other way of relabeling the elements. So, we could think of σ as having conjugacy
class
More generally, you can show that two elements of Sn are conjugate if and only if they
have the same “shape” under cycle decomposition.
Question 16.4.4.Show that the number of conjugacy classes of Snequals the number ofpartitions of n.
As long as I’ve put the above picture, I may as well also define:
Definition 16.4.5.Let G be a group. The center of G, denoted Z(G), is the set of
elements x ∈ G such that xg = gx for every g ∈ G. More succinctly,
You can check this is indeed a subgroup of G.
Question 16.4.6.Why is Z(G) normal in G?
Question 16.4.7.What are the conjugacy classes of elements in the center?
A trivial result that gets used enough that I should explicitly call it out:
Corollary 16.4.8 (Conjugacy in abelian groups is trivial) If G is abelian, then the
conjugacy classes all have size one.
16.5A few harder problems to think about
Problem 16A (PUMaC 2009 C8).Taotao wants to buy a bracelet consisting of seven beads,
each of which is orange, white or black. (The bracelet can be rotated and reflected in space.)
Find the number of possible bracelets.
Problem 16B.Show that two elements in the same conjugacy class have the same order.
Problem 16C.Prove Burnside’s lemma.
Problem 16D⋆ (The “class equation”). Let G be a finite group. We define the
centralizerCG(g) = {x ∈ G∣xg = gx} for each g ∈ G. Show that
where S ⊆ G is defined as follows: for each conjugacy class C ⊆ G with |C| > 1, we
pick a representative of C and add it to S.
Problem 16E† (Classical).Assume G is a finite group and p is the smallest
prime dividing its order. Let H be a subgroup of G with ∕ = p. Show that H is normal
in G.
17Find all groups
The following problem will hopefully never be proposed at the IMO.
Let n be a positive integer and let S = . Find all functions f : S ×S → S
such that
(a)
f(x,1) = f(1,x) = x for all x ∈ S.
(b)
f(f(x,y),z) = f(x,f(y,z)) for all x,y,z ∈ S.
(c)
For every x ∈ S there exists a y ∈ S such that f(x,y) = f(y,x) = 1.
Nonetheless, it’s remarkable how much progress we’ve made on this “problem”. In this chapter
I’ll try to talk about some things we have accomplished.
17.1Sylow theorems
Here we present the famous Sylow theorems, some of the most general results we have about
finite groups.
Theorem 17.1.1 (The Sylow theorems) Let G be a group of order pnm, where gcd(p,m) = 1
and p is a prime. A Sylow p-subgroup is a subgroup of order pn. Let np be the number of
Sylow p-subgroups of G. Then
(a)
np≡ 1 (mod p). In particular, np≠0 and a Sylow p-subgroup exists.
(b)
np divides m.
(c)
Any two Sylow p-subgroups are conjugate subgroups (hence isomorphic).
Sylow’s theorem is really huge for classifying groups; in particular, the conditions np≡ 1
(mod p) and np∣m can often pin down the value of np to just a few values. Here are some
results which follow from the Sylow theorems.
A Sylow p-subgroup is normal if and only if np = 1.
Any group G of order pq, where p < q are primes, must have nq = 1, since nq≡ 1
(mod q) yet nq∣p. Thus G has a normal subgroup of order q.
Since any abelian group has all subgroups normal, it follows that any abelian group
has exactly one Sylow p-subgroup for every p dividing its order.
If p≠q, the intersection of a Sylow p-subgroup and a Sylow q-subgroup is just
{1G}. That’s because the intersection of any two subgroups is also a subgroup, and
Lagrange’s theorem tells us that its order must divide both a power of p and a
power of q; this can only happen if the subgroup is trivial.
Here’s an example of another “practical” application.
Proposition 17.1.2 (Triple product of primes) If = pqr is the product of distinct
primes, then G must have a normal Sylow subgroup.
Proof.WLOG, assume p < q < r. Notice that np≡ 1 (mod p), np|qr and cyclically, and
assume for contradiction that np,nq,nr> 1.
Since nr|pq, we have nr = pq since nr divides neither p nor q as nr≥ 1 + r > p,q.
Also, np≥ 1 + p and nq≥ 1 + q. So we must have at least 1 + p Sylow p-subgroups, at
least 1 + q Sylow q-subgroups, and at least pq Sylow r-subgroups.
But these groups are pretty exclusive.
Question 17.1.3.Take the np+ nq+ nrSylow subgroups and consider two of them, say H1and H2. Show that = 1 as follows: check that H1∩H2is a subgroup of both H1andH2, and then use Lagrange’s theorem.
We claim that there are too many elements now. Indeed, if we count the non-identity
elements contributed by these subgroups, we get
which is more elements than G has! □
17.2(Optional) Proving Sylow’s theorem
The proof of Sylow’s theorem is somewhat involved, and in fact many proofs exist. I’ll present
one below here. It makes extensive use of group actions, so I want to recall a few facts first. If
G acts on X, then
The orbits of the action form a partition of X.
if 𝒪 is any orbit, then the orbit-Stabilizer theorem says that
for any x ∈𝒪.
In particular: suppose in the above that G is a p-group, meaning |G| = pt for some
t. Then either = 1 or p divides . In the case 𝒪 = {x}, then by definition,
x is a fixed point of every element of G: we have g ⋅ x = x for every x.
Note that when I say x is a fixed point, I mean it is fixed by every element of the group, i.e. the
orbit really has size one. Hence that’s a really strong condition.
17.2.iDefinitions
Prototypical example for this section:Conjugacy in Sn.
I’ve defined conjugacy of elements previously, but I now need to define it for groups:
Definition 17.2.1.Let G be a group, and let X denote the set of subgroups of G. Then
conjugation is the action of G on X that sends
If H and K are subgroups of G such that H = gKg−1 for some g ∈ G (in other
words, they are in the same orbit under this action), then we say they are conjugate
subgroups.
Because we somehow don’t think of conjugate elements as “that different” (for example, in
permutation groups), the following shouldn’t be surprising:
Question 17.2.2.Show that for any subgroup H of a group G, the map H → gHg−1byhghg−1is in fact an isomorphism. This implies that any two conjugate subgroups areisomorphic.
Definition 17.2.3.For any subgroup H of G the normalizer of H is defined as
In other words, it is the stabilizer of H under the conjugation action.
We are now ready to present the proof.
17.2.iiStep 1: Prove that a Sylow p-subgroup exists
What follows is something like the probabilistic method. By considering the set X of
ALL subsets of size pn at once, we can exploit the “deep number theoretic fact”
that
(It’s not actually deep: use Lucas’ theorem.)
Here is the proof.
Let G act on X by g ⋅ X : = .
Take an orbit 𝒪 with size not divisible by p. (This is possible because of our deep
number theoretic fact. Since is nonzero mod p and the orbits partition X, the
claimed orbit must exist.)
Let S ∈𝒪, H = StabG(S). Then pn divides , by the orbit-Stabilizer theorem.
Consider a second action: let H act on S by h ⋅ s : = hs (we know hs ∈ S since
H = StabG(S)).
Observe that StabH(s) = {1H}. Then all orbits of the second action must have
size . Thus divides = pn.
This implies = pn, and we’re done.
17.2.iiiStep 2: Any two Sylow p-subgroups are conjugate
Let P be a Sylow p-subgroup (which exists by the previous step). We now prove that
for any p-group Q, Q ⊆ gPg−1. Note that if Q is also a Sylow p-subgroup, then
Q = gPg−1 for size reasons; this implies that any two Sylow subgroups are indeed
conjugate.
Let Q act on the set of left cosets of P by left multiplication. Note that
Q is a p-group, so any orbit has size divisible by p unless it’s 1.
But the number of left cosets is m, which isn’t divisible by p.
Hence some coset gP is a fixed point for every q, meaning qgP = gP for all q.
Equivalently, qg ∈ gP for all q ∈ Q, so Q ⊆ gPg−1 as desired.
17.2.ivStep 3: Showing np≡ 1 (mod p)
Let 𝒮 denote the set of all the Sylow p-subgroups. Let P ∈𝒮 be arbitrary.
Question 17.2.4.Why does equal np? (In other words, are you awake?)
Now we can proceed with the proof. Let P act on 𝒮 by conjugation. Then:
Because P is a p-group, np (mod p) is the number of fixed points of this action.
Now we claim P is the only fixed point of this action.
Let Q be any other fixed point, meaning xQx−1 = Q for any x ∈ P.
Define the normalizer NG(Q) = . It contains both P and Q.
Now for the crazy part: apply Step 2 to NG(Q). Since P and Q are Sylow
p-subgroups of it, they must be conjugate.
Hence P = Q, as desired.
17.2.vStep 4: np divides m
Since np≡ 1 (mod p), it suffices to show np divides . Let G act on the set of all Sylow
p-groups by conjugation. Step 2 says this action has only one orbit, so the Orbit-Stabilizer
theorem implies np divides .
17.3(Optional) Simple groups and Jordan-Hölder
Prototypical example for this section:Decomposition of ℤ∕12ℤ is 1 ⊴ ℤ∕2ℤ ⊴ ℤ∕4ℤ ⊴ ℤ∕12ℤ.
Just like every integer breaks down as the product of primes, we can try to break every
group down as a product of “basic” groups. Armed with our idea of quotient groups, the right
notion is this.
Definition 17.3.1.A simple group is a group with no normal subgroups other than
itself and the trivial group.
Question 17.3.2.For which n is ℤ∕nℤ simple? (Hint: remember that ℤ∕nℤ is abelian.)
Then we can try to define what it means to “break down a group”.
Definition 17.3.3.A composition series of a group G is a sequence of subgroups
H0, H1, …, Hn such that
of maximal length (i.e. n is as large as possible, but all Hi are of course distinct). The
composition factors are the groups H1∕H0, H2∕H1, …, Hn∕Hn−1.
You can show that the “maximality” condition implies that the composition factors are all
simple groups.
Let’s say two composition series are equivalent if they have the same composition factors
(up to permutation); in particular they have the same length. Then it turns out that the
following theorem is true.
Theorem 17.3.4 (Jordan-Hölder) Every finite group G admits a unique composition
series up to equivalence.
Example 17.3.5 (Fundamental theorem of arithmetic when n = 12) Let’s consider the group
ℤ∕12ℤ. It’s not hard to check that the possible composition series are
{1}⊴ ℤ∕2ℤ ⊴ ℤ∕4ℤ ⊴ ℤ∕12ℤ
with factors ℤ∕2ℤ, ℤ∕2ℤ, ℤ∕3ℤ
{1}⊴ ℤ∕2ℤ ⊴ ℤ∕6ℤ ⊴ ℤ∕12ℤ
with factors ℤ∕2ℤ, ℤ∕3ℤ, ℤ∕2ℤ
{1}⊴ ℤ∕3ℤ ⊴ ℤ∕6ℤ ⊴ ℤ∕12ℤ
with factors ℤ∕3ℤ, ℤ∕2ℤ, ℤ∕2ℤ.
These correspond to the factorization 12 = 22⋅ 3.
This suggests that classifying all finite simple groups would be great progress, since every
finite group is somehow a “product” of simple groups; the only issue is that there are multiple
ways of building a group from constituents.
Amazingly, we actually have a full list of simple groups, but the list is really bizarre. Every
finite simple group falls in one of the following categories:
ℤ∕pℤ for p a prime,
For n ≥ 5, the subgroup of Sn consisting of “even” permutations.
A simple group of Lie type (which I won’t explain), and
Twenty-six “sporadic” groups which do not fit into any nice family.
The two largest of the sporadic groups have cute names. The baby monster group has
order
and the monster group (also “friendly giant”) has order
It contains twenty of the sporadic groups as subquotients (including itself), and these
twenty groups are called the “happy family”.
Math is weird.
Question 17.3.6.Show that “finite simple group of order 2” is redundant in the sense that anygroup of order 2 is both finite and simple.
17.4A few harder problems to think about
Problem 17A⋆ (Cauchy’s theorem).Let G be a group and let p be a prime dividing
. Prove1
that G has an element of order p.
Problem 17B.Let G be a finite simple group. Show that ≠56.
Problem 17C (Engel’s PSS?).Consider the set of all words consisting of
the letters a and b. Given such a word, we can change the word either by inserting
a word of the form www, where w is a word, anywhere in the given word, or by
deleting such a sequence from the word. Can we turn the word ab into the word ba?
Problem 17D.Let p be a prime. Show that the only simple group
with order pn (for some positive integer n) is the cyclic group ℤ∕pnℤ.
18The PID structure theorem
The main point of this chapter is to discuss a classification theorem for finitely generated
abelian groups. This won’t take long to do, and if you like, you can read just the first section
and then move on.
However, since I’m here, I will go ahead and state the result as a special case of the much
more general structure theorem. Its corollaries include
All finite-dimensional vector spaces are k⊕n.
The classification theorem for finitely generated abelian groups,
The Jordan decomposition of a matrix from before,
Another canonical form for a matrix: “Frobenius normal form”.
18.1Finitely generated abelian groups
Remark 18.1.1 — We talk about abelian groups in what follows, but really the morally
correct way to think about these structures is as ℤ-modules.
Definition 18.1.2.An abelian group G = (G,+) is finitely generated if it is finitely
generated as a ℤ-module. (That is, there exists a finite collection b1,…,bm∈ G, such that
every x ∈ G can be written in the form c1b1 + + cmbm for some c1,…,cm∈ ℤ.)
Example 18.1.3 (Examples of finitely generated abelian groups)
(a)
ℤ is finitely generated (by 1).
(b)
ℤ∕nℤ is finitely generated (by 1).
(c)
ℤ⊕2 is finitely generated (by two elements (1,0) and (0,1)).
(d)
ℤ⊕3⊕ ℤ∕9ℤ ⊕ ℤ∕2016ℤ is finitely generated by five elements.
(e)
ℤ∕3ℤ ⊕ ℤ∕5ℤ is finitely generated by two elements.
Exercise 18.1.4.In fact ℤ∕3ℤ ⊕ℤ∕5ℤ is generated by one element. What is it?
You might notice that these examples are not very diverse. That’s because they are actually
the only examples:
Theorem 18.1.5 (Fundamental theorem of finitely generated abelian groups) Let G be a
finitely generated abelian group. Then there exists an integer r, prime powers q1, …, qm
(not necessarily distinct) such that
This decomposition is unique up to permutation of the ℤ∕qiℤ.
Definition 18.1.6.The rank of a finitely generated abelian group G is the integer r
above.
Now, we could prove this theorem, but it is more interesting to go for the gold and state and
prove the entire structure theorem.
18.2Some ring theory prerequisites
Prototypical example for this section:R = ℤ.
Before I can state the main theorem, I need to define a few terms for UFD’s, which behave
much like ℤ:
Our intuition from the case R = ℤ basically carries over verbatim.
We don’t even need to deal with prime ideals and can factor elements instead.
Definition 18.2.1.If R is a UFD, then p ∈ R is a prime element if (p) is a prime
ideal and p≠0. For UFD’s this is equivalent to: if p = xy then either x or y is a unit.
So for example in ℤ the set of prime elements is {±2,±3,±5,…}. Now, since R is a UFD,
every element r factors into a product of prime elements
Definition 18.2.2.We say r dividess if s = r′r for some r′∈ R. This is written r∣s.
Example 18.2.3 (Divisibility in ℤ) The number 0 is divisible by every element of ℤ. All
other divisibility as expected.
Question 18.2.4.Show that r∣s if and only if the exponent of each prime in r is less than orequal to the corresponding exponent in s.
Now, the case of interest is the even stronger case when R is a PID:
Proposition 18.2.5 (PID’s are Noetherian UFD’s) If R is a PID, then it is Noetherian and
also a UFD.
Proof.The fact that R is Noetherian is obvious. For R to be a UFD we essentially repeat
the proof for ℤ, using the fact that (a,b) is principal in order to extract gcd(a,b). □
In this case, we have a Chinese remainder theorem for elements.
Theorem 18.2.6 (Chinese remainder theorem for rings) Let m and n be relatively prime
elements, meaning (m) + (n) = (1). Then
Here the ring product is as defined in ?? .
Proof.This is the same as the proof of the usual Chinese remainder theorem. First, since
(m,n) = (1) we have am + bn = 1 for some a and b. Then we have a map
One can check that this map is well-defined and an isomorphism of rings. (Diligent
readers invited to do so.) □
Finally, we need to introduce the concept of a Noetherian R-module.
Definition 18.2.7.An R-module M is Noetherian if it satisfies one of the two equivalent
conditions:
Its submodules obey the ascending chain condition: there is no infinite sequence of
modules M1⊊ M2⊊….
All submodules of M (including M itself) are finitely generated.
This generalizes the notion of a Noetherian ring: a Noetherian ring R is one for which R is
Noetherian as an R-module.
Question 18.2.8.Check these two conditions are equivalent. (Copy the proof for rings.)
18.3The structure theorem
Our structure theorem takes two forms:
Theorem 18.3.1 (Structure theorem, invariant form) Let R be a PID and let M be any
finitely generated R-module. Then
for some si (possibly zero) satisfying s1∣s2∣…∣sm.
Corollary 18.3.2 (Structure theorem, primary form) Let R be a PID and let M be any
finitely generated R-module. Then
where qi = piei for some prime element pi and integer ei≥ 1.
Proof of corollary. Factor each si into prime factors (since R is a UFD), then use the
Chinese remainder theorem. □
Remark 18.3.3 — In both theorems the decomposition is unique up to permutations
of the summands; good to know, but I won’t prove this.
18.4Reduction to maps of free R-modules
Definition 18.4.1.A free R-module is a module of the form R⊕n (or more generally,
⊕IR for some indexing set I, just to allow an infinite basis).
The proof of the structure theorem proceeds in two main steps. First, we reduce the problem
to a linear algebra problem involving free R-modules R⊕d. Once that’s done, we just have to
play with matrices; this is done in the next section.
Suppose M is finitely generated by d elements. Then there is a surjective map of
R-modules
whose image on the basis of R⊕d are the generators of M. Let K denote the kernel.
We claim that K is finitely generated as well. To this end we prove that
Lemma 18.4.2 (Direct sum of Noetherian modules is Noetherian) Let M and N be two
Noetherian R-modules. Then the direct sum M ⊕ N is also a Noetherian R-module.
Proof.It suffices to show that if L ⊆ M ⊕ N, then L is finitely generated. One guess is
that L = P ⊕Q, where P and Q are the projections of L onto M and N. Unfortunately
this is false (take M = N = ℤ and L = {(n,n)∣n ∈ ℤ}) so we will have to be more
careful.
Consider the submodules
A
= ⊆ M
B
= ⊆ N.
(Note the asymmetry for A and B: the proof doesn’t work otherwise.) Then A is finitely
generated by a1, …, ak, and B is finitely generated by b1, …, bℓ. Let xi = (ai,0) and let
yi = (∗,bi) be elements of L (where the ∗’s are arbitrary things we don’t care about). Then xi
and yi together generate L. □
Question 18.4.3.Deduce that for R a PID, R⊕dis Noetherian.
Hence K ⊆ R⊕d is finitely generated as claimed. So we can find another surjective map
R⊕f↠ K. Consequently, we have a composition
Observe that M is the cokernel of the linear map T, i.e. we have that
So it suffices to understand the map T well.
18.5Smith normal form
The idea is now that we have reduced our problem to studying linear maps T : R⊕m→ R⊕n,
which can be thought of as a generic matrix
for a basis e1, …, em of R⊕m and f1, …, fn of N.
Of course, as you might expect it ought to be possible to change the given basis of T such
that T has a nicer matrix form. We already saw this in Jordan form, where we had a map
T : V → V and changed the basis so that T was “almost diagonal”. This time, we have
two sets of bases we can change, so we would hope to get a diagonal basis, or even
better.
Before proceeding let’s think about how we might edit the matrix: what operations are
permitted? Here are some examples:
Swapping rows and columns, which just corresponds to re-ordering the basis.
Adding a multiple of a column to another column. For example, if we add 3 times
the first column to the second column, this is equivalent to replacing the basis
Adding a multiple of a row to another row. One can see that adding 3 times the
first row to the second row is equivalent to replacing the basis
More generally,
If A is an invertible n × n matrix we can replace T with AT.
This corresponds to replacing
(the “invertible” condition just guarantees the latter is a basis). Of course similarly we
can replace X with XB where B is an invertible m × m matrix; this corresponds
to
Armed with this knowledge, we can now approach:
Theorem 18.5.1 (Smith normal form) Let R be a PID. Let M = R⊕m and N = R⊕n be
free R-modules and let T : M → N be a linear map. Set k = min{m,n}.
Then we can select a pair of new bases for M and N such that T has only diagonal
entries s1, s2, …, sk and s1∣s2∣…∣sk.
So if m > n, the matrix should take the form
and similarly when m ≤ n.
Question 18.5.2.Show that Smith normal form implies the structure theorem.
Remark 18.5.3 — Note that this is not a generalization of Jordan form.
In Jordan form we consider maps T : V → V ; note that the source and target
space are the same, and we are considering one basis for the space V .
In Smith form the maps T : M → N are between different modules, and we pick
two sets of bases (one for M and one for N).
Example 18.5.4 (Example of Smith normal form) To give a flavor of the idea of the
proof, let’s work through a concrete example with the ℤ-matrix
The GCD of all the entries is 2, and so motivated by this, we perform the Euclideanalgorithm on the left column: subtract the second row from the first row, then three
times the first row from the second:
Now that the GCD of 2 is present, we move it to the upper-left by switching the two
rows, and then kill off all the entries in the same row/column; since 2 was the GCD all
along, we isolate 2 completely:
This reduces the problem to a 1 × 2 matrix. So we just apply the Euclidean algorithm
again there:
Now all we have to do is generalize this proof to work with any PID. It’s intuitively
clear how to do this: the PID condition more or less lets you perform a Euclidean
algorithm.
Proof of Smith normal form. Begin with a generic matrix
We want to show, by a series of operations (gradually changing the given basis) that
we can rearrange the matrix into Smith normal form.
Define gcd(x,y) to be any generator of the principal ideal (x,y).
Claim 18.5.5 (“Euclidean algorithm”).If a and b are entries in the same row or
column, we can change bases to replace a with gcd(a,b) and b with something else.
Proof. We do just the case of columns. By hypothesis, gcd(a,b) = xa + yb for some
x,y ∈ R. We must have (x,y) = (1) now (we’re in a UFD). So there are u and v such
that xu + yv = 1. Then
and the first matrix is invertible (check this!), as desired. ■
Let s1 = (aij)i,j be the GCD of all entries. Now by repeatedly applying this algorithm, we
can cause s to appear in the upper left hand corner. Then, we use it to kill off all the entries in
the first row and the first column, thus arriving at a matrix
Now we repeat the same procedure with this lower-right (m − 1) × (n − 1) matrix, and so
on. This gives the Smith normal form. □
With the Smith normal form, we have in the original situation that
and applying the theorem to T completes the proof of the structure theorem.
18.6A few harder problems to think about
Now, we can apply our structure theorem!
Problem 18A† (Finite-dimensional vector spaces are all isomorphic).A vector space V
over a field k has a finite spanning set of vectors. Show that Vk⊕n for some n.
Problem 18B† (Frobenius normal form).Let T : V → V where V is a
finite-dimensional vector space over an arbitrary field k (not necessarily algebraically
closed). Show that one can write T as a block-diagonal matrix whose blocks are all of
the form
(View V as a k[x]-module with action x ⋅ v = T(v).)
Problem 18C† (Jordan normal form).Let T : V → V where V is a finite-dimensional vector
space over an arbitrary field k which is algebraically closed. Prove that T can be written in
Jordan form.
Problem 18D.Find two abelian groups G and H which are not
isomorphic, but for which there are injective homomorphisms GH and HG.
Solution. Take G = ℤ∕3ℤ ⊕ ℤ∕9ℤ ⊕ ℤ∕9ℤ ⊕ ℤ∕9ℤ ⊕… and H = ℤ∕9ℤ ⊕ ℤ∕9ℤ ⊕ ℤ∕9ℤ ⊕ℤ∕9ℤ ⊕…. Then there are maps GH and HG, but the groups are not isomorphic
since e.g. G has an element g ∈ G of order 3 for which there’s no g′∈ G with g = 3g′. □
Part VI Representation Theory
19Representations of algebras
In the 19th century, the word “group” hadn’t been invented yet; all work was done with
subsets of GL(n) or Sn. Only much later was the abstract definition of a group was given, an
abstract set G which was an object in its own right.
While this abstraction is good for some reasons, it is often also useful to work
with concrete representations. This is the subject of representation theory. Linear
algebra is easier than abstract algebra, so if we can take a group G and represent it
concretely as a set of matrices in GL(n), this makes them easier to study. This is the
representation theory of groups: how can we take a group and represent its elements as
matrices?
19.1Algebras
Prototypical example for this section:k[x1,…,xn] and k[G].
Rather than working directly with groups from the beginning, it will be more convenient to
deal with so-called k-algebras. This setting is more natural and general than that of groups, so
once we develop the theory of algebras well enough, it will be fairly painless to specialize to
the case of groups.
Colloquially,
An associative k-algebra is a possibly noncommutative ring with a copy of k insideit. It is thus a k-vector space.
I’ll present examples before the definition:
Example 19.1.1 (Examples of k-Algebras) Let k be any field. The following are examples of
k-algebras:
(a)
The field k itself.
(b)
The polynomial ring k[x1,…,xn].
(c)
The set of n × n matrices with entries in k, which we denote by Matn(k). Note the
multiplication here is not commutative.
(d)
The set Mat(V ) of linear operators T : V → V , with multiplication given by the
composition of operators. (Here V is some vector space over k.) This is really the
same as the previous example.
Definition 19.1.2.Let k be a field. A k-algebraA is a possibly noncommutative ring,
equipped with an injective ring homomorphism kA (whose image is the “copy of k”).
In particular, 1k1A.
Thus we can consider k as a subset of A, and we then additionally require λ⋅a = a⋅λ
for each λ ∈ k and a ∈ A.
If the multiplication operation is also commutative, then we say A is a commutativealgebra.
Definition 19.1.3.Equivalently, a k-algebraA is a k-vector space which also has an
associative, bilinear multiplication operation (with an identity 1A). The “copy of k” is
obtained by considering elements λ1A for each λ ∈ k (i.e. scaling the identity by the
elements of k, taking advantage of the vector space structure).
Abuse of Notation 19.1.4.Some other authors don’t require A to be associative or
to have an identity, so to them what we have just defined is an “associative algebra with
1”. However, this is needlessly wordy for our purposes.
Example 19.1.5 (Group algebra) The group algebrak[G] is the k-vector space whose
basis elements are the elements of a group G, and where the product of two basis elements
is the group multiplication. For example, suppose G = ℤ∕2ℤ = {1G,x}. Then
with multiplication given by
Question 19.1.6.When is k[G] commutative?
The example k[G] is very important, because (as we will soon see) a representation of the
algebra k[G] amounts to a representation of the group G itself.
It is worth mentioning at this point that:
Definition 19.1.7.A homomorphism of k-algebras A, B is a linear map T : A → B
which respects multiplication (i.e. T(xy) = T(x)T(y)) and which sends 1A to 1B. In
other words, T is both a homomorphism as a ring and as a vector space.
Definition 19.1.8.Given k-algebras A and B, the direct sumA ⊕ B is defined as
pairs a + b, where addition is done in the obvious way, but we declare ab = 0 for any
a ∈ A and b ∈ B.
Question 19.1.9.Show that 1A+ 1Bis the multiplicative identity of A ⊕ B.
19.2Representations
Prototypical example for this section:k[S3] acting on k⊕3is my favorite.
Definition 19.2.1.A representation of a k-algebra A (also a left A-module)
is:
(i)
A k-vector space V , and
(ii)
An action ⋅ of A on V : thus, for every a ∈ A we can take v ∈ V and act on it to get a⋅v.
This satisfies the usual axioms:
(a + b) ⋅ v = a ⋅ v + b ⋅ v, a ⋅ (v + w) = a ⋅ v + a ⋅ w, and (ab) ⋅ v = a ⋅ (b ⋅ v).
λ ⋅ v = λv for λ ∈ k. In particular, 1A⋅ v = v.
Definition 19.2.2.The action of A can be more succinctly described as saying that
there is a k-algebra homomorphism ρ : A → Mat(V ). (So a ⋅ v = ρ(a)(v).) Thus we can
also define a representation of A as a pair
This is completely analogous to how a group action G on a set X with n elements just
amounts to a group homomorphism G → Sn. From this perspective, what we are really trying
to do is:
If A is an algebra, we are trying to represent the elements of A as matrices.
Abuse of Notation 19.2.3.While a representation is a pair (V,ρ) of both the vector
space V and the action ρ, we frequently will just abbreviate it to “V ”. This is probably
one of the worst abuses I will commit, but everyone else does it and I fear the mob.
Abuse of Notation 19.2.4.Rather than ρ(a)(v) we will just write ρ(a)v.
Example 19.2.5 (Representations of Mat(V ))
(a)
Let A = Mat2(ℝ). Then there is a representation (ℝ⊕2,ρ) where a matrix a ∈ A
just acts by a ⋅ v = ρ(a)(v) = a(v).
(b)
More generally, given a vector space V over any field k, there is an obvious
representation of A = Mat(V ) by a ⋅ v = ρ(a)(v) = a(v) (since a ∈ Mat(V )).
From the matrix perspective: if A = Mat(V ), then we can just represent A as
matrices over V .
(c)
There are other representations of A = Mat2(ℝ). A silly example is the
representation (ℝ⊕4,ρ) given by
More abstractly, viewing ℝ⊕4 as (ℝ⊕2) ⊕ (ℝ⊕2), this is a ⋅ (v1,v2) = (a ⋅ v1,a ⋅ v2).
Example 19.2.6 (Representations of polynomial algebras)
(a)
Let A = k. Then a representation of k is just any k-vector space V .
(b)
If A = k[x], then a representation (V,ρ) of A amounts to a vector space V plus the
choice of a linear operator T ∈ Mat(V ) (by T = ρ(x)).
(c)
If A = k[x]∕(x2) then a representation (V,ρ) of A amounts to a vector space V plus
the choice of a linear operator T ∈ Mat(V ) satisfying T2 = 0.
(d)
We can create arbitrary “functional equations” with this pattern. For example, if
A = k[x,y]∕(x2− x + y,y4) then representing A by V amounts to finding operators
S,T ∈ Mat(V ) satisfying S2 = S − T and T4 = 0.
Example 19.2.7 (Representations of groups)
(a)
Let A = ℝ[S3]. Then let
We can let A act on V as follows: given a permutation π ∈ S3, we permute the
corresponding coordinates in V . So for example, if
This extends linearly to let A act on V , by permuting the coordinates.
From the matrix perspective, what we are doing is representing the permutations in
S3 as permutation matrices on k⊕3, like
(b)
More generally, let A = k[G]. Then a representation (V,ρ) of A amounts to a group
homomorphism ψ : G → GL(V ). (In particular, ρ(1G) = idV.) We call this a grouprepresentation of G.
Example 19.2.8 (Regular representation) Any k-algebra A is a representation (A,ρ)
over itself, with a ⋅ b = ρ(a)(b) = ab (i.e. multiplication given by A). This is called the
regular representation, denoted Reg(A).
19.3Direct sums
Prototypical example for this section:The example with ℝ[S3] seems best.
Definition 19.3.1.Let A be k-algebra and let V = (V,ρV) and W = (W,ρW) be two
representations of A. Then V ⊕ W is a representation, with action ρ given by
This representation is called the direct sum of V and W.
Example 19.3.2 Earlier we let Mat2(ℝ) act on ℝ⊕4 by
So this is just a direct sum of two two-dimensional representations.
More generally, given representations (V,ρV) and (W,ρW) the representation ρ of V ⊕ W
looks like
Example 19.3.3 (Representation of Sn decomposes) Let A = ℝ[S3] again, acting via
permutation of coordinates on
Consider the two subspaces
W1
=
W2
= .
Note V = W1⊕ W2 as vector spaces. But each of W1 and W2 is a subrepresentation
(since the action of A keeps each Wi in place), so V = W1⊕ W2 as representations
too.
Direct sums also come up when we play with algebras.
Proposition 19.3.4 (Representations of A ⊕ B are VA⊕ VB) Let A and B be k-algebras.
Then every representation of A ⊕ B is of the form
where VA and VB are representations of A and B, respectively.
Sketch of Proof. Let (V,ρ) be a representation of A⊕B. For any v ∈ V , ρ(1A +1B)v =
ρ(1A)v + ρ(1B)v. One can then set VA = {ρ(1A)v∣v ∈ V } and VB = {ρ(1B)v∣v ∈ V }.
These are disjoint, since if ρ(1A)v = ρ(1B)v′, we have ρ(1A)v = ρ(1A1A)v = ρ(1A1B)v′ =
0V, and similarly for the other side. □
19.4Irreducible and indecomposable representations
Prototypical example for this section:k[S3] decomposes as the sum of two spaces.
One of the goals of representation theory will be to classify all possible representations of an
algebra A. If we want to have a hope of doing this, then we want to discard “silly”
representations such as
and focus our attention instead on “irreducible” representations. This motivates:
Definition 19.4.1.Let V be a representation of A. A subrepresentationW ⊆ V is
a subspace W with the property that for any a ∈ A and w ∈ W, a ⋅ w ∈ W. In other
words, this subspace is invariant under actions by A.
Thus for example if V = W1⊕ W2 for representations W1, W2 then W1 and W2 are
subrepresentations of V .
Definition 19.4.2.If V has no proper nonzero subrepresentations then it is
irreducible. If there is no pair of proper subrepresentations W1, W2 such that
V = W1⊕ W2, then we say V is indecomposable.
Definition 19.4.3.For brevity, an irrep of an algebra/group is a finite-dimensional
irreducible representation.
Example 19.4.4 (Representation of Sn decomposes) Let A = ℝ[S3] again, acting via
permutation of coordinates on
Consider again the two subspaces
W1
=
W2
= .
As we’ve seen, V = W1⊕W2, and thus V is not irreducible. But one can show that W1 and
W2 are irreducible (and hence indecomposable) as follows.
For W1 it’s obvious, since W1 is one-dimensional.
For W2, consider any vector w = (a,b,c) with a + b + c = 0 and not all zero. Then
WLOG we can assume a≠b (since not all three coordinates are equal). In that case,
(1 2) sends w to w′ = (b,a,c). Then w and w′ span W2.
Thus V breaks down completely into irreps.
Unfortunately, if W is a subrepresentation of V , then it is not necessarily the case that we
can find a supplementary vector space W′ such that V = W ⊕ W′. Put another way, if V is
reducible, we know that it has a subrepresentation, but a decomposition requires two
subrepresentations. Here is a standard counterexample:
Exercise 19.4.5.Let A = ℝ[x], and V = ℝ⊕2be the representation with action
Show that the only subrepresentation is W = {(t,0)∣t ∈ ℝ}. So V is not irreducible, but it isindecomposable.
Here is a slightly more optimistic example, and the “prototypical example” that you should
keep in mind.
Exercise 19.4.6.Let A =Matd(k) and consider the obvious representation k⊕dof A that wedescribed earlier. Show that it is irreducible. (This is obvious if you understand the definitionswell enough.)
19.5Morphisms of representations
We now proceed to define the morphisms between representations.
Definition 19.5.1.Let (V,ρV) and (W,ρW) be representations of A. An intertwiningoperator, or morphism, is a linear map T : V → W such that
for any a ∈ A, v ∈ V . (Note that the first ⋅ is the action of ρV and the second ⋅ is
the action of ρW.) This is exactly what you expect if you think that V and W are “left
A-modules”. If T is invertible, then it is an isomorphism of representations and we say
VW.
Remark 19.5.2 (For commutative diagram lovers) — The condition T(a⋅v) = a⋅T(v) can be
read as saying that
commutes for any a ∈ A.
Remark 19.5.3 (For category lovers) — A representation is just a “bilinear” functor
from an abelian one-object category {∗} (so Hom(∗,∗)A) to the abelian category
Vectk. Then an intertwining operator is just a natural transformation.
Here are some examples of intertwining operators.
Example 19.5.4 (Intertwining operators)
(a)
For any λ ∈ k, the scalar map T(v) = λv is intertwining.
(b)
If W ⊆ V is a subrepresentation, then the inclusion WV is an intertwining
operator.
(c)
The projection map V1⊕ V2↠ V1 is an intertwining operator.
(d)
Let V = ℝ⊕2 and represent A = k[x] by (V,ρ) where
Thus ρ(x) is rotation by 90∘ around the origin. Let T be rotation by 30∘. Then
T : V → V is intertwining (the rotations commute).
Exercise 19.5.5 (Kernel and image are subrepresentations).Let T : V → W be an intertwiningoperator.
(a)
Show thatkerT ⊆ V is a subrepresentation of V .
(b)
Show thatimT ⊆ W is a subrepresentation of W.
The previous lemma gives us the famous Schur’s lemma.
Theorem 19.5.6 (Schur’s lemma) Let V and W be representations of a k-algebra A. Let
T : V → W be a nonzero intertwining operator. Then
(a)
If V is irreducible, then T is injective.
(b)
If W is irreducible, then T is surjective.
In particular if both V and W are irreducible then T is an isomorphism.
An important special case is if k is algebraically closed: then the only intertwining operators
T : V → V are multiplication by a constant.
Theorem 19.5.7 (Schur’s lemma for algebraically closed fields) Let k be an algebraically
closed field. Let V be an irrep of a k-algebra A. Then any intertwining operator
T : V → V is multiplication by a scalar.
Exercise 19.5.8.Use the fact that T has an eigenvalue λ to deduce this from Schur’s lemma.(Consider T − λ ⋅idV, and use Schur to deduce it’s zero.)
We have already seen the counterexample of rotation by 90∘ for k = ℝ; this was the same
counterexample we gave to the assertion that all linear maps have eigenvalues.
19.6The representations of Matd(k)
To give an example of the kind of progress already possible, we prove:
Theorem 19.6.1 (Representations of Matd(k)) Let k be any field, d be a positive integer
and let W = k⊕d be the obvious representation of A = Matd(k). Then the only
finite-dimensional representations of Matd(k) are W⊕n for some positive integer n (up
to isomorphism). In particular, it is irreducible if and only if n = 1.
For concreteness, I’ll just sketch the case d = 2, since the same proof applies verbatim to
other situations. This shows that the examples of representations of Mat2(ℝ) we gave earlier
are the only ones.
As we’ve said this is essentially a functional equation. The algebra A = Mat2(k) has basis
given by four matrices
satisfying relations like E1 + E2 = idA, Ei2 = Ei, E1E2 = 0, etc. So let V be a
representation of A, and let Mi = ρ(Ei) for each i; we want to classify the possible
matrices Mi on V satisfying the same functional equations. This is because, for
example,
By the same token M1M3 = M3. Proceeding in a similar way, we can obtain the following
multiplication table:
Note that each Mi is a linear operator V → V ; for all we know, it could have hundreds of
entries. Nonetheless, given the multiplication table of the basis Ei we get the corresponding
table for the Mi.
So, in short, the problem is as follows:
Find all vector spaces V and quadruples of matrices Misatisfying themultiplication table above.
Let W1 = M1img(V ) and W2 = M2img(V ) be the images of M1 and M2.
Claim 19.6.2.V = W1⊕ W2.
Proof.First, note that for any v ∈ V we have
Moreover, we have that W1∩ W2 = {0}, because if M1v1 = M2v2 then M1v1 =
M1(M1v1) = M1(M2v2) = 0. □
Claim 19.6.3.W1W2.
Proof.Check that the maps
are well-defined and mutually inverse. □
Now, let e1,…,en be basis elements of W1; thus M4e1, …, M4en are basis elements of W2.
However, each {ej,M4ej} forms a basis of a subrepresentation isomorphic to W = k⊕2 (what’s
the isomorphism?).
This finally implies that all representations of A are of the form W⊕n. In particular, W is
irreducible because there are no representations of smaller dimension at all!
19.7A few harder problems to think about
Problem 19A†. Suppose we have one-dimensional representations V1 = (V1,ρ1) and
V2 = (V2,ρ2) of A. Show that V1V2 if and only if ρ1(a) and ρ2(a) are multiplication
by the same constant for every a ∈ A.
Problem 19B† (Schur’s lemma for commutative algebras).Let A be a commutative algebra
over an algebraically closed field k. Prove that any irrep of A is one-dimensional.
Problem 19C⋆. Let (V,ρ) be a representation of A. Then Mat(V ) is a representation
of A with action given by
for T ∈ Mat(V ).
(a)
Show that ρ : Reg(A) → Mat(V ) is an intertwining operator.
(b)
If V is d-dimensional, show that Mat(V )V⊕d as representations of A.
Problem 19D⋆. Fix an algebra A. Find all intertwining operators
Problem 19E.Let (V,ρ) be an indecomposable (not irreducible)
representation of an algebra A. Prove that any intertwining operator T : V → V is either
nilpotent or an isomorphism.
(Note that ?? doesn’t apply, since the field k may not be algebraically closed.)
20Semisimple algebras
In what follows, assume the field k is algebraically closed.
Fix an algebra A and suppose you want to study its representations. We have a “direct
sum” operation already. So, much like we pay special attention to prime numbers, we’re
motivated to study irreducible representations and then build all the representations of A from
there.
Unfortunately, we have seen (?? ) that there exists a representation which is not
irreducible, and yet cannot be broken down as a direct sum (indecomposable). This is weirdand bad, so we want to give a name to representations which are more well-behaved. We
say that a representation is completely reducible if it doesn’t exhibit this bad
behavior.
Even better, we say a finite-dimensional algebra A is semisimple if all its finite-dimensional
representations are completely reducible. So when we study finite-dimensional representations
of semisimple algebras A, we just have to figure out what the irreps are, and then piecing them
together will give all the representations of A.
In fact, semisimple algebras A have even nicer properties. The culminating point of the
chapter is when we prove that A is semisimple if and only if A⊕iMat(Vi), where the Vi
are the irreps of A (yes, there are only finitely many!).
20.1Schur’s lemma continued
Prototypical example for this section:For V irreducible, Homrep(V⊕2,V⊕2)k⊕4.
Definition 20.1.1.For an algebra A and representations V and W, we let
Homrep(V,W) be the set of intertwining operators between them. (It is also a k-algebra.)
By Schur’s lemma (since k is algebraically closed, which again, we are taking as a standing
assumption), we already know that if V and W are irreps, then
Can we say anything more? For example, it also tells us that
The possible maps are v(c1v1,c2v2) for some choice of c1,c2∈ k.
More generally, suppose V is an irrep and consider Homrep(V⊕m,V⊕n). Intertwining
operators are determined completely T : V⊕m→ V⊕n by the mn choices of compositions
where the first arrow is inclusion to the ith component of V⊕m (for 1 ≤ i ≤ m)
and the second arrow is inclusion to the jth component of V⊕n (for 1 ≤ j ≤ n).
However, by Schur’s lemma on each of these compositions, we know they must be
constant.
Thus, Homrep(V⊕n,V⊕m) consist of n×m “matrices” of constants, and the map is provided
by
where the cij∈ k but vi∈ V ; note the type mismatch! This is not just a linear map
V⊕ni→ V⊕mi; rather, the outputs are m linear combinations of the inputs.
More generally, we have:
Theorem 20.1.2 (Schur’s lemma for completely reducible representations) Let V and W
be completely reducible representations, and set V = ⊕V i⊕ni, W = ⊕V i⊕mi for
integers ni,mi≥ 0, where each Vi is an irrep. Then
meaning that an intertwining operator T : V → W amounts to, for each i, an ni× mi
matrix of constants which gives a map V i⊕ni→ V i⊕mi.
Corollary 20.1.3 (Subrepresentations of completely reducible representations) Let V =
⊕V i⊕ni be completely reducible. Then any subrepresentation W of V is isomorphic to
⊕V i⊕mi where mi≤ ni for each i, and the inclusion WV is given by the direct sum
of inclusion V i⊕miV i⊕ni, which are ni× mi matrices.
Proof.Apply Schur’s lemma to the inclusion WV . □
20.2Density theorem
We are going to take advantage of the previous result to prove that finite-dimensional algebras
have finitely many irreps.
Theorem 20.2.1 (Jacobson density theorem) Let (V1,ρ1), …, (Vr,ρr) be pairwise
nonisomorphic finite-dimensional representations of A. Then there is a surjective map of
vector spaces
The right way to think about this theorem is that
Density is the “Chinese remainder theorem” for irreps of A.
Recall that in number theory, the Chinese remainder theorem tells us that given lots
of “unrelated” congruences, we can find a single N which simultaneously satisfies
them all. Similarly, given lots of different nonisomorphic representations of A, this
means that we can select a single a ∈ A which induces any tuple (ρ1(a),…,ρr(a)) of
actions we want — a surprising result, since even the r = 1 case is not obvious at
all!
This also gives us the non-obvious corollary
Corollary 20.2.2 (Finiteness of number of representations) Any finite-dimensional algebra
A has at most dimA irreps.
Proof.If Vi are such irreps then A ↠⊕iV i⊕dimVi, hence we have the inequality
∑
(dimVi)2≤ dimA. □
Proof of density theorem. Let V = V1⊕⊕Vr, so A acts on V = (V,ρ) by ρ = ⊕iρi.
Thus by ?? , we can instead consider ρ as an intertwining operator
We will use this instead as it will be easier to work with.
First, we handle the case r = 1. Fix a basis e1, …, en of V = V1. Assuming for contradiction
that the map is not surjective. Then there is a map of representations (by ρ and
the isomorphism) Reg(A) → V⊕n given by a(a ⋅ e1,…,a ⋅ en). By hypothesis is
not surjective: its image is a proper subrepresentation of V⊕n. Assume its image is
isomorphic to V⊕m for m < n, so by ?? there is a matrix of constants X with
where the two arrows in the top row have the same image; hence the pre-image (v1,…,vm) of
(e1,…,en) can be found. But since r < n we can find constants c1,…,cn not all zero such that
X applied to the column vector (c1,…,cn) is zero:
contradicting the fact that ei are linearly independent. Hence we conclude the theorem for
r = 1.
As for r ≥ 2, the image ρimg(A) is necessarily of the form ⊕iV i⊕ri (by ?? ) and by the
above ri = dimVi for each i. □
20.3Semisimple algebras
Definition 20.3.1.A finite-dimensional algebra A is a semisimple if every
finite-dimensional representation of A is completely reducible.
Theorem 20.3.2 (Semisimple algebras) Let A be a finite-dimensional algebra. Then the
following are equivalent:
(i)
A⊕iMatdi(k) for some di.
(ii)
A is semisimple.
(iii)
Reg(A) is completely reducible.
Proof.(i) (ii) follows from ?? and ?? . (ii) (iii) is tautological.
To see (iii) (i), we use the following clever trick. Consider
On one hand, by ?? , it is isomorphic to Aop (A with opposite multiplication), because
the only intertwining operators Reg(A) → Reg(A) are those of the form −⋅ a. On the
other hand, suppose that we have set Reg(A) = ⊕iV i⊕ni. By ?? , we have
But Matn(k)opMatn(k) (just by transposing), so we recover the desired
conclusion. □
In fact, if we combine the above result with the density theorem (and ?? ), we
obtain:
Theorem 20.3.3 (Sum of squares formula) For a finite-dimensional algebra A we have
where the Vi are the irreps of A; equality holds exactly when A is semisimple, in which
case
Proof.The inequality was already mentioned in ?? . It is equality if and only if the map
ρ : A →⊕iMat(Vi) is an isomorphism; this means all Vi are present. □
Remark 20.3.4 (Digression) — For any finite-dimensional A, the kernel of the map
ρ : A →⊕iMat(Vi) is denoted Rad(A) and is the so-called Jacobson radical of A;
it’s the set of all a ∈ A which act by zero in all irreps of A. The usual definition of
“semisimple” given in books is that this Jacobson radical is trivial.
20.4Maschke’s theorem
We now prove that the representation theory of groups is as nice as possible.
Theorem 20.4.1 (Maschke’s theorem) Let G be a finite group, and k an algebraically
closed field whose characteristic does not divide |G|. Then k[G] is semisimple.
This tells us that when studying representations of groups, all representations are
completely reducible.
Proof.Consider any finite-dimensional representation (V,ρ) of k[G]. Given a proper
subrepresentation W ⊆ V , our goal is to construct a supplementary G-invariant subspace
W′ which satisfies
This will show that indecomposable ⇔ irreducible, which is enough to show k[G] is
semisimple.
Let π : V → W be any projection of V onto W, meaning π(v) = v⇔v ∈ W. We
consider the averaging map P : V → V by
We’ll use the following properties of the map:
Exercise 20.4.2.Show that the map P satisfies:
For any w ∈ W, P(w) = w.
For any v ∈ V , P(w) ∈ W.
The map P : V → V is an intertwining operator.
Thus P is idempotent (it is the identity on its image W), so by ?? we have
V = kerP ⊕ imP, but both kerP and imP are subrepresentations as desired. □
Remark 20.4.3 — In the case where k = ℂ, there is a shorter proof. Suppose B :
V × V → ℂ is an arbitrary bilinear form. Then we can “average” it to obtain a new
bilinear form
The averaged form is G-invariant, in the sense that = . Then,
one sees that if W ⊆ V is a subrepresentation, so is its orthogonal complement W⊥.
This implies the result.
20.5Example: the representations of ℂ[S3]
We compute all irreps of ℂ[S3]. I’ll take for granted right now there are exactly three such
representations (which will be immediate by the first theorem in the next chapter: we’ll in fact
see that the number of representations of G is exactly equal to the number of conjugacy
classes of G).
Given that, if the three representations of have dimension d1, d2, d3 , then we ought to
have
From this, combined with some deep arithmetic, we deduce that we should have d1 = d2 = 1
and d3 = 2 or some permutation.
In fact, we can describe these representations explicitly. First, we define:
Definition 20.5.1.Let G be a group. The complex trivial group representation of
a group G is the one-dimensional representation ℂtriv = (ℂ,ρ) where g ⋅ v = v for all
g ∈ G and v ∈ ℂ (i.e. ρ(g) = id for all g ∈ G).
Remark 20.5.2 (Warning) — The trivial representation of an algebra A doesn’t make
sense for us: we might want to set a ⋅ v = v but this isn’t linear in A. (You could try to
force it to work by deleting the condition 1A⋅ v = v from our definition; then one can
just set a ⋅ v = 0. But even then ℂtriv would not be the trivial representation of k[G].)
Then the representations are:
The one-dimensional ℂtriv; each σ ∈ S3 acts by the identity.
There is a nontrivial one-dimensional representation ℂsign where the map S3→ ℂ×
is given by sending σ to the sign of σ. Thus in ℂsign every σ ∈ S3 acts as ±1. Of
course, ℂtriv and ℂsign are not isomorphic (as one-dimensional representations are
never isomorphic unless the constants they act on coincide for all a, as we saw in
?? ).
Finally, we have already seen the two-dimensional representation, but now
we give it a name. Define refl0 to be the representation whose vector space
is {(x,y,z)∣x + y + z = 0}, and whose action of S3 on it is permutation of
coordinates.
Exercise 20.5.3.Show thatrefl0is irreducible, for example by showing directly that nosubspace is invariant under the action of S3.
Thus V is also not isomorphic to the previous two representations.
This implies that these are all the irreps of S3. Note that, if we take the representation V of S3
on k⊕3, we just get that V = refl0⊕ ℂtriv.
20.6A few harder problems to think about
Problem 20A.Find all the irreps of ℂ[ℤ∕nℤ].
Problem 20B (Maschke requires |G| finite).Consider the representation of the group
ℝ on ℂ⊕2 under addition by a homomorphism
Show that this representation is not irreducible, but it is indecomposable.
Problem 20C.Prove that all irreducible representations of a finite group are finite-dimensional.
Problem 20D.Determine all the complex irreps of D10.
21Characters
Characters are basically the best thing ever. To every representation V of A we will attach a
so-called character χV : A → k. It will turn out that the characters of irreps of V will
determine the representation V completely. Thus an irrep is just specified by a set of dimA
numbers.
21.1Definitions
Definition 21.1.1.Let V = (V,ρ) be a finite-dimensional representation of A. The
characterχV : A → k attached to A is defined χV = Tr∘ρ, i.e.
Since Tr and ρ are additive, this is a k-linear map (but it is not multiplicative). Note also that
χV ⊕W = χV + χW for any representations V and W.
We are especially interested in the case A = k[G], of course. As usual, we just have
to specify χV(g) for each g ∈ S3 to get the whole map k[G] → k. Thus we often
think of χV as a function G → k, called a character of the group G. Here is the case
G = S3:
Example 21.1.2 (Character table of S3) Let’s consider the three irreps of G = S3 from
before. For ℂtriv all traces are 1; for ℂsign the traces are ±1 depending on sign
(obviously, for one-dimensional maps k → k the trace “is” just the map itself). For refl0
we take a basis (1,0,−1) and (0,1,−1), say, and compute the traces directly in this
basis.
The above table is called the character table of the group G. The table above has certain
mysterious properties, which we will prove as the chapter progresses.
(I)
The value of χV(g) only depends on the conjugacy class of g.
(II)
The number of rows equals the number of conjugacy classes.
(III)
The sum of the squares of any row is 6 again!
(IV)
The “dot product” of any two rows is zero.
Abuse of Notation 21.1.3.The name “character” for χV : G → k is a bit of a
misnomer. This χV is not multiplicative in any way, as the above example shows: one
can almost think of it as an element of k⊕|G|.
Question 21.1.4.Show that χV(1A) =dimV , so one can read the dimensions of therepresentations from the leftmost column of a character table.
21.2The dual space modulo the commutator
For any algebra, we first observe that since Tr(TS) = Tr(ST), we have for any V
that
This explains observation (I) from earlier:
Question 21.2.1.Deduce that if g and h are in the same conjugacy class of a group G, and Vis a representation of ℂ[G], then χ(g) = χ(h).
Now, given our algebra A we define the commutator [A,A] to be the (two-sided)
ideal1
generated by elements of the form xy −yx. Thus [A,A] is contained in the kernel of each χV.
Definition 21.2.2.The space A∕[A,A] is called the abelianization of A; for brevity
we denote it as Aab. We think of this as “A modulo the relation ab = ba for each
a,b ∈ A.”
So we can think of each character χV as an element of (Aab)∨
.
Example 21.2.3 (Examples of abelianizations)
(a)
If A is commutative, then [A,A] = {0} and Aab=A.
(b)
If A = Matk(d), then [A,A] consists exactly of the d × d matrices of trace zero.
(Proof: harmless exercise.) Consequently, Aab is one-dimensional.
(c)
Suppose A = k[G]. We claim that dimAab is equal to the number of conjugacy
classes of A. Indeed, an element of A can be thought of as just an arbitrary function
ξ : G → k. So an element of Aab is a function ξ : G → k such that ξ(gh) = ξ(hg)
for every g,h ∈ G. This is equivalent to functions from conjugacy classes of G to k.
Theorem 21.2.4 (Character of representations of algebras) Let A be an algebra over an
algebraically closed field. Then
(a)
Characters of pairwise non-isomorphic irreps are linearly independent as elements
of Aab.
(b)
If A is finite-dimensional and semisimple, then the characters attached to irreps form
a basis of Aab.
In particular, in (b) the number of irreps of A equals dimAab.
Proof.Part (a) is more or less obvious by the density theorem. Suppose there is a linear
dependence, so that for every a we have
for some integer r.
Question 21.2.5.Deduce that c1== cr= 0 from the density theorem.
For part (b), assume there are r irreps we may assume that
where V1, …, Vr are the irreps of A. Since we have already showed the characters are
linearly independent we need only show that dim(A∕[A,A]) = r, which follows from the
observation earlier that each Mat(Vi) has a one-dimensional abelianization. □
Since G has dim ℂ[G]ab conjugacy classes, this completes the proof of (II).
21.3Orthogonality of characters
Now we specialize to the case of finite groups G, represented over ℂ.
Definition 21.3.1.Let Classes(G) denote the set conjugacy classes of G.
If G has r conjugacy classes, then it has r irreps. Each (finite-dimensional) representation V ,
irreducible or not, gives a character χV.
Abuse of Notation 21.3.2.From now on, we will often regard χV as a function
G → ℂ or as a function Classes(G) → ℂ. So for example, we will write both χV(g)
(for g ∈ G) and χV(C) (for a conjugacy class C); the latter just means χV(gC) for any
representative gC∈ C.
Definition 21.3.3.Let Funclass(G) denote the set of functions Classes(G) → ℂ viewed
as a vector space over ℂ. We endow it with the inner form
This is the same “dot product” that we mentioned at the beginning, when we looked at the
character table of S3. We now aim to prove the following orthogonality theorem, which will
imply (III) and (IV) from earlier.
Theorem 21.3.4 (Orthogonality) For any finite-dimensional complex representations V
and W of G we have
In particular, if V and W are irreps then
Corollary 21.3.5 (Irreps give an orthonormal basis) The characters associated to irreps
form an orthonormal basis of Funclass(G).
In order to prove this theorem, we have to define the dual representation and the tensor
representation, which give a natural way to deal with the quantity χV(g)χW(g).
Definition 21.3.6.Let V = (V,ρ) be a representation of G. The dual representationV∨ is the representation on V∨ with the action of G given as follows: for each ξ ∈ V∨,
the action of g gives a g ⋅ ξ ∈ V∨ specified by
Definition 21.3.7.Let V = (V,ρV) and W = (W,ρW) be group representations of G.
The tensor product of V and W is the group representation on V ⊗W with the action
of G given on pure tensors by
which extends linearly to define the action of G on all of V ⊗ W.
Remark 21.3.8 — Warning: the definition for tensors does not extend to algebras. We
might hope that a ⋅ (v ⊗ w) = (a ⋅ v) ⊗ (a ⋅ w) would work, but this is not even linear in
a ∈ A (what happens if we take a = 2, for example?).
Theorem 21.3.9 (Character traces) If V and W are finite-dimensional representations of G,
then for any g ∈ G.
(a)
χV ⊕W(g) = χV(g) + χW(g).
(b)
χV ⊗W(g) = χV(g) ⋅ χW(g).
(c)
χV∨(g) =χV(g).
Proof.Parts (a) and (b) follow from the identities Tr(S ⊕ T) = Tr(S) + Tr(T) and
Tr(S ⊗T) = Tr(S)Tr(T). However, part (c) is trickier. As (ρ(g))|G| = ρ(g|G|) = ρ(1G) =
idV by Lagrange’s theorem, we can diagonalize ρ(g), say with eigenvalues λ1, …, λn which
are |G|th roots of unity, corresponding to eigenvectors e1, …, en. Then we see that in the
basis e1∨, …, en∨, the action of g on V∨ has eigenvalues λ1−1, λ2−1, …, λn−1. So
where the last step follows from the identity |z| = 1 ⇔z−1 =z. □
Remark 21.3.10 (Warning) — The identities (b) and (c) do not extend linearly to ℂ[G],
i.e. it is not true for example that χV(a) =χV(a) if we think of χV as a map ℂ[G] → ℂ.
Proof of orthogonality relation. The key point is that we can now reduce the sums of
products to just a single character by
So we can rewrite the sum in question as just
Let P : V ⊗ W∨→ V ⊗ W∨ be the action of ∑g∈Gg, so we wish to find TrP.
Exercise 21.3.11.Show that P is idempotent. (Compute P ∘ P directly.)
Hence V ⊗ W∨ = kerP ⊕ imP (by ?? ) and imP is the subspace of elements which are
fixed under G. From this we deduce that
Now, consider the natural isomorphism V ⊗ W∨→ Hom(W,V ).
Exercise 21.3.12.Let g ∈ G. Show that under this isomorphism, T ∈Hom(W,V ) satisfiesg ⋅T = T if and only if T(g ⋅w) = g ⋅T(w) for each w ∈ W. (This is just unwinding three or fourdefinitions.)
The orthogonality relation gives us a fast and mechanical way to check whether a
finite-dimensional representation V is irreducible. Namely, compute the traces χV(g) for
each g ∈ G, and then check whether = 1. So, for example, we could have
seen the three representations of S3 that we found were irreps directly from the
character table. Thus, we can now efficiently verify any time we have a complete set of
irreps.
21.4Examples of character tables
Example 21.4.1 (Dihedral group on 10 elements) Let D10 = .
Let ω = exp(). We write four representations of D10:
ℂtriv, all elements of D10 act as the identity.
ℂsign, r acts as the identity while s acts by negation.
V1, which is two-dimensional and given by r and s.
V2, which is two-dimensional and given by r and s.
We claim that these four representations are irreducible and pairwise non-isomorphic. We do
so by writing the character table:
Then a direct computation shows the orthogonality relations, hence we indeed have an
orthonormal basis. For example, = 1 + 2 ⋅ 1 + 2 ⋅ 1 + 5 ⋅ (−1) = 0.
Example 21.4.2 (Character table of S4) We now have enough machinery to to compute
the character table of S4, which has five conjugacy classes (corresponding to cycle types
id, 2, 3, 4 and 2+2). First of all, we note that it has two one-dimensional representations,
ℂtriv and ℂsign, and these are the only ones (because there are only two homomorphisms
S4→ ℂ×). So thus far we have the table
Now, the remaining three representations have dimensions d1, d2, d3 with
which has only (d1,d2,d3) = (2,3,3) and permutations. Now, we can take the refl0
representation
with basis (1,0,0,−1), (0,1,0,−1) and (0,0,1,−1). This can be geometrically checked to
be irreducible, but we can also do this numerically by computing the character directly
(this is tedious): it comes out to have 3, 1, 0, −1, −1 which indeed gives norm
Note that we can also tensor this with the sign representation, to get another irreducible
representation (since ℂsign has all traces ±1, the norm doesn’t change). Finally, we
recover the final row using orthogonality (which we name ℂ2, for lack of a better name);
hence the completed table is as follows.
21.5A few harder problems to think about
Problem 21A† (Reading decompositions from characters).Let W be a complex
representation of a finite group G. Let V1, …, Vr be the complex irreps of G and set
ni = . Prove that each ni is a non-negative integer and
Problem 21B.Consider complex representations of G = S4. The representation refl0⊗ refl0
is 9-dimensional, so it is clearly reducible. Compute its decomposition in terms of the five
irreducible representations.
Problem 21C (Tensoring by one-dimensional irreps).Let V and W be irreps of G, with
dimW = 1. Show that V ⊗ W is irreducible.
Problem 21D (Quaternions).Compute the character table of the quaternion group Q8.
Problem 21E⋆ (Second orthogonality formula).Let g and h be
elements of a finite group G, and let V1, …, Vr be the irreps of G. Prove that
Here, CG(g) = is the centralizer of g.
22Some applications
With all this setup, we now take the time to develop some nice results which are of
independent interest.
22.1Frobenius divisibility
Theorem 22.1.1 (Frobenius divisibility) Let V be a complex irrep of a finite group G.
Then dimV divides |G|.
The proof of this will require algebraic integers (developed in the algebraic number theory
chapter). Recall that an algebraic integer is a complex number which is the root of a monic
polynomial with integer coefficients, and that these algebraic integers form a ring ℤ under
addition and multiplication, and that ℤ∩ ℚ = ℤ.
First, we prove:
Lemma 22.1.2 (Elements of ℤ[G] are integral) Let α ∈ ℤ[G]. Then there exists a monic
polynomial P with integer coefficients such that P(α) = 0.
Proof.Let Ak be the ℤ-span of 1,α1,…,αk. Since ℤ[G] is Noetherian, the inclusions
A0⊆ A1⊆ A2⊆… cannot all be strict, hence Ak = Ak+1 for some k, which means αk+1
can be expressed in terms of lower powers of α. □
Proof of Frobenius divisibility. Let C1, …, Cm denote the conjugacy classes of G. Then
consider the rational number
we will show it is an algebraic integer, which will prove the theorem. Observe that we
can rewrite it as
We split the sum over conjugacy classes, so
We claim that for every i,
is an algebraic integer, where
To see this, note that Ti commutes with elements of G, and hence is an intertwining
operator Ti : V → V . Thus by Schur’s lemma, Ti = λi⋅ idV and TrT = λi dimV . By
?? , λi∈ℤ, as desired.
Now we are done, since χV(Ci)∈ℤ too (it is the sum of conjugates of roots of unity),
so is the sum of products of algebraic integers, hence itself an algebraic integer. □
22.2Burnside’s theorem
We now prove a group-theoretic result. This is the famous poster child for representation
theory (in the same way that RSA is the poster child of number theory) because the result is
purely group theoretic.
Recall that a group is simple if it has no normal subgroups. In fact, we will prove:
Theorem 22.2.1 (Burnside) Let G be a nonabelian group of order paqb (where p,q are
distinct primes and a,b ≥ 0). Then G is not simple.
In what follows p and q will always denote prime numbers.
Lemma 22.2.2 (On gcd(|C|,dimV ) = 1) Let V = (V,ρ) be an complex irrep of G. Assume
C is a conjugacy class of G with gcd(|C|,dimV ) = 1. Then for any g ∈ C, either
ρ(g) is multiplication by a scalar, or
χV(g) = Trρ(g) = 0.
Proof.If 𝜀i are the n eigenvalues of ρ(g) (which are roots of unity), then from the proof
of Frobenius divisibility we know χV(g) ∈ℤ, thus from gcd(|C|,n) = 1 we get
So this follows readily from a fact from algebraic number theory, namely ?? : either
𝜀1 = = 𝜀n (first case) or 𝜀1 + + 𝜀n = 0 (second case). □
Lemma 22.2.3 (Simple groups don’t have prime power conjugacy classes) Let G be a finite
simple group. Then G cannot have a conjugacy class of order pk (where k > 0).
Proof.By contradiction. Assume C is such a conjugacy class, and fix any g ∈ C. By
the second orthogonality formula (?? ) applied g and 1G (which are not conjugate since
g≠1G) we have
where Vi are as usual all irreps of G.
Exercise 22.2.4.Show that there exists a nontrivial irrep V such that p ∤dimV and χV(g)≠0.(Proceed by contradiction to show that −∈ℤif not.)
Let V = (V,ρ) be the irrep mentioned. By the previous lemma, we now know that ρ(g) acts
as a scalar in V .
Now consider the subgroup
We claim this is a nontrivial normal subgroup of G. It is easy to check H is normal, and
since |C| > 1 we have that H is nontrivial. As represented by V each element of H acts
trivially in G, so since V is nontrivial and irreducible, H≠G. This contradicts the assumption
that G was simple. □
With this lemma, Burnside’s theorem follows by partitioning the |G| elements of our group
into conjugacy classes. Assume for contradiction G is simple. Each conjugacy class must
have order either 1 (of which there are |Z(G)| by ?? ) or divisible by pq (by the
previous lemma), but on the other hand the sum equals |G| = paqb. Consequently,
we must have |Z(G)| > 1. But G is not abelian, hence Z(G)≠G, thus the center
Z(G) is a nontrivial normal subgroup, contradicting the assumption that G was
simple.
22.3Frobenius determinant
We finish with the following result, the problem that started the branch of representation
theory. Given a finite group G, we create n variables {xg}g∈G, and an n×n matrix MG whose
(g,h)th entry is xgh.
Example 22.3.1 (Frobenius determinants)
(a)
If G = ℤ∕2ℤ = then the matrix would be
Then detMG = (xid− xT)(xid + xT).
(b)
If G = S3, a long computation gives the irreducible factorization of detMG is
where F(a,b,c) = a2 + b2 + c2− ab − bc − ca; the latter factor is irreducible.
Theorem 22.3.2 (Frobenius determinant) The polynomial detMG (in |G| variables) factors into
a product of irreducible polynomials such that
(i)
The number of polynomials equals the number of conjugacy classes of G, and
(ii)
The multiplicity of each polynomial equals its degree.
You may already be able to guess how the “sum of squares” result is related! (Indeed, look
at deg detMG.)
Legend has it that Dedekind observed this behavior first in 1896. He didn’t know how to
prove it in general, so he sent it in a letter to Frobenius, who created representation theory to
solve the problem.
With all the tools we’ve built, it is now fairly straightforward to prove the result.
Proof.Let V = (V,ρ) = Reg(ℂ[G]) and let V1, …, Vr be the irreps of G. Let’s consider
the map T : ℂ[G] → ℂ[G] which has matrix MG in the usual basis of ℂ[G], namely
Thus we want to examine detT.
But we know that V = ⊕i=1rV i⊕dimVi as before, and so breaking down T over its
subspaces we know
So we only have to show two things: the polynomials detTVi are irreducible, and they
are pairwise different for different i.
Let Vi = (Vi,ρ), and pick k = dimVi.
Irreducible: By the density theorem, for any M ∈ Mat(Vi) there exists a particular
choice of complex numbers xg∈ G such that
View ρi(g) as a k × k matrix with complex coefficients. Thus the “generic”
(T↾Vi)({xg}), viewed as a matrix with polynomial entries, must have linearly
independent entries (or there would be some matrix in Mat(Vi) that we can’t
achieve).
Then, the assertion follows (by a linear variable change) from the simple fact that
the polynomial det(yij)1≤i,j≤m in m2 variables is always irreducible.
Pairwise distinct: We show that from detT|Vi({xg}) we can read off the character χVi,
which proves the claim. In fact
Exercise 22.3.3.Pick any basis for Vi. IfdimVi= k, and 1G≠g ∈ G, then
Thus, we are done. □
Part VII Quantum Algorithms
23Quantum states and measurements
In this chapter we’ll explain how to set up quantum states using linear algebra. This will allow
me to talk about quantum circuits in the next chapter, which will set the stage for Shor’s
algorithm.
I won’t do very much physics (read: none at all). That is, I’ll only state what the physical
reality is in terms of linear algebras, and defer the philosophy of why this is true to your
neighborhood “Philosophy of Quantum Mechanics” class (which is a “social science” class at
MIT!).
23.1Bra-ket notation
Physicists have their own notation for vectors: whereas I previously used something like v, e1,
and so on, in this chapter you’ll see the infamous bra-ket notation: a vector will be denoted
by , where ∙ is some variable name: unlike in math or Python, this can include
numbers, symbols, Unicode characters, whatever you like. This is called a “ket”.
To pay a homage to physicists everywhere, we’ll use this notation for this chapter
too.
Abuse of Notation 23.1.1 (For this part, dimH < ∞).In this part on quantum
computation, we’ll use the word “Hilbert space” as defined earlier, but in fact all our
Hilbert spaces will be finite-dimensional.
If dimH = n, then its orthonormal basis elements are often denoted
(instead of ei) and a generic element of H denoted by
and various other Greek letters.
Now for any ∈ H, we can consider the canonical dual element in H∨ (since H has an
inner form), which we denote by ⟨ψ| (a “bra”). For example, if dimH = 2 then we can
write
in an orthonormal basis, in which case
We even can write dot products succinctly in this notation: if = , then the dot
product of and is given by
So we will use the notation instead of the more mathematical . In
particular, the squared norm of is just . Concretely, for dimH = 2 we have
= |α|2 + |β|2.
23.2The state space
If you think that’s weird, well, it gets worse.
In classical computation, a bit is either 0 or 1. More generally, we can think of a classical
space of n possible states 0, …, n − 1. Thus in the classical situation, the space of possible
states is just a discrete set with n elements.
In quantum computation, a qubit is instead any complex linear combination of 0 and 1. To
be precise, consider the normed complex vector space
and denote the orthonormal basis elements by and . Then a qubit is a nonzero
element ∈ H, so that it can be written in the form
where α and β are not both zero. Typically, we normalize so that has norm
1:
In particular, we can recover the “classical” situation with ∈ H and ∈ H, but now
we have some “intermediate” states, such as
Philosophically, what has happened is that:
Instead of allowing just the states and , we allow any complex linearcombination of them.
More generally, if dimH = n, then the possible states are nonzero elements
which we usually normalize so that |c0|2 + |c1|2 + + |cn−1|2 = 1.
23.3Observations
Prototypical example for this section: id corresponds to not making a measurement sinceall its eigenvalues are equal, but any operator with distinct eigenvalues will causecollapse.
If you think that’s weird, well, it gets worse. First, some linear algebra:
Definition 23.3.1.Let V be a finite-dimensional inner product space. For a map T : V → V ,
the following conditions are equivalent:
= for any x,y ∈ V .
T = T†.
A map T satisfying these conditions is called Hermitian.
Question 23.3.2.Show that T is normal.
Thus, we know that T is diagonalizable with respect to the inner form, so for a suitable
basis we can write it in an orthonormal basis as
As we’ve said, this is fantastic: not only do we have a basis of eigenvectors, but the
eigenvectors are pairwise orthogonal, and so they form an orthonormal basis of
V .
Question 23.3.3.Show that all eigenvalues of T are real. (T = T†.)
Back to quantum computation. Suppose we have a state ∈ H, where dimH = 2; we
haven’t distinguished a particular basis yet, so we just have a nonzero vector. Then the way
observations work (and this is physics, so you’ll have to take my word for it) is as
follows:
Pick a Hermitian operator T : H → H; then observations of T return eigenvalues ofT.
To be precise:
Pick a Hermitian operator T : H → H, which is called the observable.
Consider its eigenvalues λ0, …, λn−1 and corresponding eigenvectors T, …,
T. Tacitly we may assume that T, …, T form an orthonormal
basis of H. (The subscript T is here to distinguish the eigenvectors of T from the
basis elements of H.)
Write in the orthonormal basis as
Then the probability of observing λi is
This is called making an observation along T.
Note that in particular, for any nonzero constant c, and c are indistinguishable, which is
why we like to normalize . But the queerest thing of all is what happens to : by
measuring it, we actually destroy information. This behavior is called quantum collapse.
Suppose for simplicity that we observe with T and obtain an eigenvalue λ, and
that T is the only eigenvector with this eigenvalue. Then, the state collapses
to just the state ciT: all the other information is destroyed. (In fact, we may as
well say it collapses to T, since again constant factors are not relevant.)
More generally, if we observe λ, consider the generalized eigenspace Hλ (i.e. the
span of eigenvectors with the same eigenvalue). Then the physical state has
been changed as well: it has now been projected onto the eigenspace Hλ. In still
other words, after observation, the state collapses to
In other words,
When we make a measurement, the coefficients from different eigenspaces aredestroyed.
Why does this happen? Beats me…physics (and hence real life) is weird. But anyways, an
example.
Example 23.3.4 (Quantum measurement of a state ) Let H = ℂ⊕2 with
orthonormal basis and and consider the state
(a)
Let
This has eigenvectors = T and = T, with eigenvalues +1 and −1. So
if we measure to T, we get +1 with probability 1∕5 and −1 with probability
4∕5. After this measurement, the original state collapses to if we measured +1,
and if we measured −1. So we never learn the original probabilities.
(b)
Now consider T = id, and arbitrarily pick two orthonormal eigenvectors T, T;
thus ψ = c0T + c1T. Since all eigenvalues of T are +1, our measurement will
always be +1 no matter what we do. But there is also no collapsing, because none
of the coefficients get destroyed.
(c)
Now consider
The two normalized eigenvectors are
with eigenvalues +7 and −7 respectively. In this basis, we have
So we get +7 with probability and −7 with probability , and after the
measurement, collapses to one of T and T.
Question 23.3.5.Suppose we measure with T and get λ. What happens if we measurewith T again?
For H = ℂ⊕2 we can come up with more classes of examples using the so-called Paulimatrices. These are the three Hermitian matrices
These matrices are important because:
Question 23.3.6.Show that these three matrices, plus the identity matrix, form a basis for theset of Hermitian 2 × 2 matrices.
So the Pauli matrices are a natural choice of basis.
Their normalized eigenvectors are
which we call “z-up”, “z-down”, “x-up”, “x-down”, “y-up”, “y-down”. (The eigenvalues
are +1 for “up” and −1 for “down”.) So, given a state ∈ ℂ⊕2 we can make a
measurement with respect to any of these three bases by using the corresponding Pauli
matrix.
In light of this, the previous examples were (a) measuring along σz, (b) measuring along id,
and (c) measuring along 7σx.
Notice that if we are given a state , and are told in advance that it is either or
(or any other orthogonal states) then we are in what is more or less a classical situation.
Specifically, if we make a measurement along σx, then we find out which state that was in
(with 100% certainty), and the state does not undergo any collapse. Thus, orthogonal states
are reliably distinguishable.
23.4Entanglement
Prototypical example for this section:Singlet state: spooky action at a distance.
If you think that’s weird, well, it gets worse.
Qubits don’t just act independently: they can talk to each other by means of a tensorproduct. Explicitly, consider
endowed with the norm described in ?? . One should think of this as a qubit A in a space
HA along with a second qubit B in a different space HB, which have been allowed to
interact in some way, and H = HA⊗ HB is the set of possible states of both qubits.
Thus
is an orthonormal basis of H; here A is the basis of the first ℂ⊕2 while B is the basis
of the second ℂ⊕2, so these vectors should be thought of as “unrelated” just as with any tensor
product. The pure tensors mean exactly what you want: for example A⊗B means “0
for qubit A and 1 for qubit B”.
As before, a measurement of a state in H requires a Hermitian map H → H.
In particular, if we only want to measure the qubit B along MB, we can use the
operator
The eigenvalues of this operator coincide with the ones for MB, and the eigenspace
for λ will be the HA⊗ (HB)λ, so when we take the projection the A qubit will be
unaffected.
This does what you would hope for pure tensors in H:
Example 23.4.1 (Two non-entangled qubits) Suppose we have qubit A in the state
A + A and qubit B in the state B + B. So, the two qubits in
tandem are represented by the pure tensor
Suppose we measure along
The eigenspace decomposition is
+1 for the span of A⊗B and A⊗B, and
−1 for the span of A⊗B and A⊗B.
(We could have used other bases, like A⊗B and A⊗B for the first
eigenspace, but it doesn’t matter.) Expanding in the four-element basis, we find that we’ll
get the first eigenspace with probability
and the second eigenspace with probability as well. (Note how the coefficients for A don’t do
anything!) After the measurement, we destroy the coefficients of the other eigenspace; thus
(after re-normalization) we obtain the collapsed state
again with 50% probability each.
So this model lets us more or less work with the two qubits independently: when we make
the measurement, we just make sure to not touch the other qubit (which corresponds to the
identity operator).
Exercise 23.4.2.Show that ifidA⊗ σxBis applied to the in this example, there is nocollapse at all. What’s the result of this measurement?
Since the ⊗ is getting cumbersome to write, we say:
Abuse of Notation 23.4.3.From now on A⊗B will be abbreviated to just
, and similarly for , , .
Example 23.4.4 (Simultaneously measuring a general 2-Qubit state) Consider a
normalized state in H = ℂ⊕2⊗ ℂ⊕2, say
We can make a measurement along the diagonal matrix T : H → H with
Thus we get each of the eigenvalues 0, 1, 2, 3 with probability |α|2, |β|2, |γ|2, |δ|2. So if
we like we can make “simultaneous” measurements on two qubits in the same way that
we make measurements on one qubit.
However, some states behave very weirdly.
Example 23.4.5 (The singlet state) Consider the state
which is called the singlet state. One can see that is not a simple tensor, which
means that it doesn’t just consist of two qubits side by side: the qubits in HA and HB
have become entangled.
Now, what happens if we measure just the qubit A? This corresponds to making the
measurement
The eigenspace decomposition of T can be described as:
The span of and , with eigenvalue +1.
The span of and , with eigenvalue −1.
So one of two things will happen:
With probability , we measure +1 and the collapsed state is .
With probability , we measure −1 and the collapsed state is .
But now we see that measurement along A has told us what the state of the bit B is
completely!
By solely looking at measurements on A, we learn B; this paradox is called spooky action ata distance, or in Einstein’s tongue, spukhafte Fernwirkung. Thus,
In tensor products of Hilbert spaces, states which are not pure tensors correspondto “entangled” states.
What this really means is that the qubits cannot be described independently; the state of
the system must be given as a whole. That’s what entangled states mean: the qubits somehow
depend on each other.
23.5A few harder problems to think about
Problem 23A.We measure by σxA⊗ idB, and hence obtain either +1 or −1.
Determine the state of qubit B from this measurement.
Problem 23B (Greenberger-Horne-Zeilinger paradox).Consider the state in (ℂ⊕2)⊗3
Find the value of the measurements along each of
As for the paradox: what happens if you multiply all these measurements together?
24Quantum circuits
Now that we’ve discussed qubits, we can talk about how to use them in circuits. The key
change — and the reason that quantum circuits can do things that classical circuits cannot —
is the fact that we are allowing linear combinations of 0 and 1.
24.1Classical logic gates
In classical logic, we build circuits which take in some bits for input, and output some more
bits for input. These circuits are built out of individual logic gates. For example, the ANDgate can be pictured as follows.
One can also represent the AND gate using the “truth table”:
Similarly, we have the OR gate and the NOT gate:
We also have a so-called COPY gate, which duplicates a bit.
Of course, the first theorem you learn about these gates is that:
Theorem 24.1.1 (AND, OR, NOT, COPY are universal) The set of four gates AND, OR,
NOT, COPY is universal in the sense that any boolean function f : {0,1}n→{0,1} can
be implemented as a circuit using only these gates.
Proof.Somewhat silly: we essentially write down a circuit that OR’s across all input
strings in fpre(1). For example, suppose we have n = 3 and want to simulate the function
f(abc) with f(011) = f(110) = 1 and 0 otherwise. Then the corresponding Boolean
expression for f is simply
Clearly, one can do the same for any other f, and implement this logic into a circuit. □
Remark 24.1.2 — Since x and y = not ((not x) or (not y)), it follows that in fact, we
can dispense with the AND gate.
24.2Reversible classical logic
Prototypical example for this section:CNOT gate, Toffoli gate.
For the purposes of quantum mechanics, this is not enough. To carry through the analogy
we in fact need gates that are reversible, meaning the gates are bijections from the input
space to the output space. In particular, such gates must take the same number of input and
output gates.
Example 24.2.1 (Reversible gates)
(a)
None of the gates AND, OR, COPY are reversible for dimension reasons.
(b)
The NOT gate, however, is reversible: it is a bijection {0,1}→{0,1}.
Example 24.2.2 (The CNOT gate) The controlled-NOT gate, or the CNOT gate, is a
reversible 2-bit gate with the following truth table.
In other words, this gate XOR’s the first bit to the second bit, while leaving the first bit
unchanged. It is depicted as follows.
The first dot is called the “control”, while the ⊕ is the “negation” operation: the first
bit controls whether the second bit gets flipped or not. Thus, a typical application might
be as follows.
So, NOT and CNOT are the only nontrivial reversible gates on two bits.
We now need a different definition of universal for our reversible gates.
Definition 24.2.3.A set of reversible gates can simulate a Boolean function f(x1…xn), if
one can implement a circuit which takes
As input, x1…xn plus some fixed bits set to 0 or 1, called ancilla bits1 .
As output, the input bits x1,…,xn, the output bit f(x1,…,xn), and possibly some
extra bits (called garbage bits).
The gate(s) are universal if they can simulate any Boolean function.
For example, the CNOT gate can simulate the NOT gate, using a single ancilla bit 1,
according to the following circuit.
Unfortunately, it is not universal.
Proposition 24.2.4 (CNOT⇏AND) The CNOT gate cannot simulate the boolean
function “x and y”.
Sketch of Proof. One can see that any function simulated using only CNOT gates must
be of the form
because CNOT is the map (x,y)(x,x+y). Thus, even with ancilla bits, we can only
create functions of the form ax + by + c (mod 2) for fixed a, b, c. The AND gate is not
of this form. □
So, we need at least a three-qubit gate. The most commonly used one is:
Definition 24.2.5.The three-bit Toffoli gate, also called the CCNOT gate, is given
by
So the Toffoli has two controls, and toggles the last bit if and only if both of the
control bits are 1.
This replacement is sufficient.
Theorem 24.2.6 (Toffoli gate is universal) The Toffoli gate is universal.
Proof.We will show it can reversibly simulate AND, NOT, hence OR, which we know
is enough to show universality. (We don’t need COPY because of reversibility.)
For the AND gate, we draw the circuit
with one ancilla bit, and no garbage bits.
For the NOT gate, we use two ancilla 1 bits and one garbage bit:
This completes the proof. □
Hence, in theory we can create any classical circuit we desire using the Toffoli gate alone. Of
course, this could require exponentially many gates for even the simplest of functions.
Fortunately, this is NO BIG DEAL because I’m a math major, and having 2n gates is a
problem best left for the CS majors.
24.3Quantum logic gates
In quantum mechanics, since we can have linear combinations of basis elements, our logic
gates will instead consist of linear maps. Moreover, in quantum computation, gates are always
reversible, which was why we took the time in the previous section to show that we can
still simulate any function when restricted to reversible gates (e.g. using the Toffoli
gate).
First, some linear algebra:
Definition 24.3.1.Let V be a finite dimensional inner product space. Then for a map
U : V → V , the following are equivalent:
= for x,y ∈ V .
U† is the inverse of U.
= for x ∈ V .
The map U is called unitary if it satisfies these equivalent conditions.
Then
Quantum logic gates are unitary matrices.
In particular, unlike the classical situation, quantum gates are always reversible (and hence
they always take the same number of input and output bits).
For example, consider the CNOT gate. Its quantum analog should be a unitary map
UCNOT : H → H, where H = ℂ⊕2⊗ ℂ⊕2, given on basis elements by
So pictorially, the quantum CNOT gate is given by
OK, so what? The whole point of quantum mechanics is that we allow linear qubits to be in
linear combinations of and , too, and this will produce interesting results. For example,
let’s take = (−) and plug it into the top, with on the bottom, and see what
happens:
which is the fully entangled singlet state! Picture:
Thus, when we input mixed states into our quantum gates, the outputs are often entangled
states, even when the original inputs are not entangled.
Example 24.3.2 (More examples of quantum gates)
(a)
Every reversible classical gate that we encountered before has a quantum analog
obtained in the same way as CNOT: by specifying the values on basis elements. For
example, there is a quantum Tofolli gate which for example sends
(b)
The Hadamard gate on one qubit is a rotation given by
Thus, it sends to and to . Note that the Hadamard gate is its own
inverse. It is depicted by an “H” box.
(c)
More generally, if U is a 2 × 2 unitary matrix (i.e. a map ℂ⊕2→ ℂ⊕2) then there is
U-rotation gate similar to the previous one, which applies U to the input.
For example, the classical NOT gate is represented by U = σx.
(d)
A controlled U-rotation gate generalizes the CNOT gate. Let U : ℂ⊕2→ ℂ⊕2 be
a rotation gate, and let H = ℂ⊕2⊗ ℂ⊕2 be a 2-qubit space. Then the controlled U
gate has the following circuit diagrams.
Thus, U is applied when the controlling bit is 1, and CNOT is the special case
U = σx. As before, we get interesting behavior if the control is mixed.
And now, some more counterintuitive quantum behavior. Suppose we try to use CNOT as a
copy, with truth table.
The point of this gate is to be used with a garbage 0 at the bottom to try and simulate a
“copy” operation. So indeed, one can check that
Thus we can copy and . But as we’ve already seen if we input ⊗ into U, we
end up with the entangled state which is decisively not the ⊗ we wanted. And
in fact, the so-called no-cloning theorem implies that it’s impossible to duplicate an
arbitrary ; the best we can do is copy specific orthogonal states as in the classical case. See
also ?? .
24.4Deutsch-Jozsa algorithm
The Deutsch-Jozsa algorithm is the first example of a nontrivial quantum algorithm which
cannot be performed classically: it is a “proof of concept” that would later inspire Grover’s
search algorithm and Shor’s factoring algorithm.
The problem is as follows: we’re given a function f : {0,1}n→{0,1}, and promised that the
function f is either
A constant function, or
A balanced function, meaning that exactly half the inputs map to 0 and half the
inputs map to 1.
The function f is given in the form of a reversible black box Uf which is the control of a NOT
gate, so it can be represented as the circuit diagram
i.e. if f(x1,…,xn) = 0 then the gate does nothing, otherwise the gate flips the y bit at the
bottom. The slash with the n indicates that the top of the input really consists of n
qubits, not just the one qubit drawn, and so the black box Uf is a map on n + 1
qubits.
The problem is to determine, with as few calls to the black box Uf as possible, whether f is
balanced or constant.
Question 24.4.1.Classically, show that in the worst case we may need up to 2n−1+ 1 calls tothe function f to answer the question.
So with only classical tools, it would take O(2n) queries to determine whether f is balanced
or constant. However,
Theorem 24.4.2 (Deutsch-Jozsa) The Deutsch-Jozsa problem can be determined in a
quantum circuit with only a single call to the black box.
Proof.For concreteness, we do the case n = 1 explicitly; the general case is contained
in ?? . We claim that the necessary circuit is
Here the H’s are Hadamard gates, and the meter at the end of the rightmost wire
indicates that we make a measurement along the usual , basis. This is not a typo!
Even though classically the top wire is just a repeat of the input information, we are
about to see that it’s the top we want to measure.
Note that after the two Hadamard operations, the state we get is
H⊗2↦−−−→⊗
= ⊗− + ⊗−.
So after applying Uf, we obtain
where the modulo 2 has been left implicit. Now, observe that the effect of going
from − to − is merely to either keep the state the
same (if f(x) = 0) or to negate it (if f(x) = 1). So we can simplify and factor to
get
Thus, the picture so far is:
In particular, the resulting state is not entangled, and we can simply discard the last qubit
(!). Now observe:
If f is constant, then the upper-most state is ±.
If f is balanced, then the upper-most state is ±.
So simply doing a measurement along σx will give us the answer. Equivalently, perform another
H gate (so that H = , H = ) and measuring along σz in the usual ,
basis. Thus for n = 1 we only need a single call to the oracle. □
24.5A few harder problems to think about
Problem 24A (Fredkin gate).The Fredkin gate (also called the controlled swap, or
CSWAP gate) is the three-bit gate with the following truth table:
Thus the gate swaps the last two input bits whenever the first bit is 1. Show that this gate
is also reversible and universal.
Problem 24B (Baby no-cloning theorem). Show that there is no unitary map U on
two qubits which sends U(⊗) = ⊗ for any qubit , i.e. the following
circuit diagram is impossible.
Problem 24C (Deutsch-Jozsa). Given the black box Uf described in the
Deutsch-Jozsa algorithm, consider the following circuit.
That is, take n copies of , apply the Hadamard rotation to all of them, apply Uf,
reverse the Hadamard to all n input bits (again discarding the last bit), then measure
all n bits in the / basis (as in ?? ).
Show that the probability of measuring is 1 if f is constant and 0 if f is balanced.
Problem 24D† (Barenco et al, 1995; arXiv:quant-ph/9503016v1).Let
Verify that the quantum Toffoli gate can be implemented using just controlled
rotations via the circuit
This was a big surprise to researchers when discovered, because classical reversible logic
requires three-bit gates (e.g. Toffoli, Fredkind).
25Shor’s algorithm
OK, now for Shor’s Algorithm: how to factor M = pq in O time.
25.1The classical (inverse) Fourier transform
The “crux move” in Shor’s algorithm is the so-called quantum Fourier transform. The Fourier
transform is used to extract periodicity in data, and it turns out the quantum analogue is a lot
faster than the classical one.
Let me throw the definition at you first. Let N be a positive integer, and let ωN = exp.
Definition 25.1.1.Given a tuple of complex numbers
its discrete inverse Fourier transform is the sequence (y0,y1,…,yN−1) defined by
Equivalently, one is applying the matrix
The reason this operation is important is because it lets us detect if the xi are
periodic:
Example 25.1.2 (Example of discrete inverse Fourier transform) Let N = 6, ω = ω6 = exp()
and suppose (x0,x1,x2,x3,x4,x5) = (0,1,0,1,0,1) (hence xi is periodic modulo 2).
Thus,
y0
= = 1∕2
y1
= = 0
y2
= = 0
y3
= = −1∕2
y4
= = 0
y5
= = 0.
Thus, in the inverse transformation the “amplitudes” are all concentrated at multiples of 3;
thus this reveals the periodicity of the original sequence by = 2.
More generally, given a sequence of 1’s appearing with period r, the amplitudes will peak at
inputs which are divisible by .
Remark 25.1.3 — The fact that this operation is called the “inverse” Fourier
transform is mostly a historical accident (as my understanding goes). Confusingly, the
corresponding quantum operation is the (not-inverted) Fourier transform.
If we apply the definition as written, computing the transform takes O(N2) time. It turns
out that by an algorithm called the fast Fourier transform (whose details we won’t discuss),
one can reduce this to O(N log N) time. However, for Shor’s algorithm this is also insufficient;
we need something like O instead. This is where the quantum Fourier transform
comes in.
25.2The quantum Fourier transform
Note that to compute a Fourier transform, we need to multiply an N × N matrix with an
N-vector, so this takes O(N2) multiplications. However, we are about to show that with a
quantum computer, one can do this using O((log N)2) quantum gates when N = 2n, on a
system with n qubits.
First, some more notation:
Abuse of Notation 25.2.1.In what follows, will refer to ⊗⊗⊗
where x = xnxn−1…x1 in binary. For example, if n = 3 then really means
⊗⊗.
Observe that the n-qubit space now has an orthonormal basis , , …,
Definition 25.2.2.Consider an n-qubit state
The quantum Fourier transform is defined by
In other words, using the basis , …, , UQFT is given by the matrix
This is the exactly the same definition as before, except we have a factor added so that
UQFT is unitary. But the trick is that in the quantum setup, the matrix can be
rewritten:
Proposition 25.2.3 (Tensor representation) Let = . Then
UQFT() =
⊗
⊗…
⊗
Proof.Direct (and quite annoying) computation. In short, expand everything. □
So by using mixed states, we can deal with the quantum Fourier transform using this
“multiplication by tensor product” trick that isn’t possible classically.
Now, without further ado, here’s the circuit. Define the rotation matrices
Then, for n = 3 the circuit is given by by using controlled Rk’s as follows:
Exercise 25.2.4.Show that in this circuit, the image of of (for binary xi) is
as claimed.
For general n, we can write this as inductively as
Question 25.2.5.Convince yourself that when n = 3 the two circuits displayed are equivalent.
Thus, the quantum Fourier transform is achievable with O(n2) gates, which is enormously
better than the O(N log N) operations achieved by the classical fast Fourier transform (where
N = 2n).
25.3Shor’s algorithm
The quantum Fourier transform is the key piece of Shor’s algorithm. Now that we have it, we
can solve the factoring problem.
Let p,q > 3 be odd primes, and assume p≠q. The main idea is to turn factoring an integer
M = pq into a problem about finding the order of x (mod M); the latter is a “periodicity”
problem that the quantum Fourier transform will let us solve. Specifically, say that an x
(mod M) is good if
(i)
gcd(x,M) = 1,
(ii)
The order r of x (mod M) is even, and
(iii)
Factoring 0 ≡ (xr∕2− 1)(xr∕2 + 1) (mod M), neither of the two factors is 0
(mod M). Thus one of them is divisible by p, and the other is divisible by q.
Exercise 25.3.1 (For contest number theory practice).Show that for M = pq at least half ofthe residues in (ℤ∕Mℤ)×are good.
So if we can find the order of an arbitrary x ∈ (ℤ∕Mℤ)×, then we just keep picking x until
we pick a good one (this happens more than half the time); once we do, we compute
gcd(xr∕2− 1,M) using the Euclidean algorithm to extract one of the prime factors of M, and
we’re home free.
Now how do we do this? The idea is not so difficult: first we generate a sequence which is
periodic modulo r.
Example 25.3.2 (Factoring 77: generating the periodic state) Let’s say we’re trying to
factor M = 77, and we randomly select x = 2, and want to find its order r. Let n = 13
and N = 213, and start by initializing the state
Now, build a circuit Ux (depending on x = 2!) which takes to .
Applying this to ⊗ gives
Now suppose we measure the second qubit, and get a state of . That tells us that
the collapsed state now, up to scaling, is
The bottleneck is actually the circuit Ux; one can compute xk (mod M) by using repeated
squaring, but it’s still the clumsy part of the whole operation.
In general, the operation is:
Pick a sufficiently large N = 2n (say, N ≥ M2).
Generate = ∑k=02n−1.
Build a circuit Ux which computes .
Apply it to get a state ∑k=02n−1⊗.
Measure the second qubit to cause the first qubit to collapse to something which
is periodic modulo r. Let denote the left qubit.
Suppose we apply the quantum Fourier transform to the left qubit now: since the left
bit is periodic modulo r, we expect the transform will tell us what r is. Unfortunately,
this doesn’t quite work out, since N is a power of two, but we don’t expect r to
be.
Nevertheless, consider a state
so for example previously we had k0 = 7 if we measured 128 on x = 2. Applying the
quantum Fourier transform, we see that the coefficient of in the transformed image is
equal to
As this is a sum of roots of unity, we realize we have destructive interference unless
ωNjr = 1 (since N is large). In other words, we approximately have
up to scaling as usual. The bottom line is that
If we measure UQFTwe obtain a such thatis close to an s ∈ ℤ.
And thus given sufficient luck we can use continued fractions to extract the value of
r.
Example 25.3.3 (Finishing the factoring of M = 77) As
before, we made an observation to the second qubit, and thus the first qubit collapses to
the state = + + …. Now we make a measurement and obtain j = 4642,
which means that for some integer s we have
Now, we analyze the continued fraction of ; we find the first few convergents are
So is a very good approximation, hence we deduce s = 17 and r = 30 as candidates.
And indeed, one can check that r = 30 is the desired order.
This won’t work all the time (for example, we could get unlucky and measure j = 0,
i.e. s = 0, which would tell us no information at all).
But one can show that we succeed any time that
This happens at least of the time, and since r < M this means that given
sufficiently many trials, we will eventually extract the correct order r. This is Shor’s
algorithm.
Part VIII Calculus 101
26Limits and series
Now that we have developed the theory of metric (and topological) spaces well,
we give a three-chapter sequence which briskly covers the theory of single-variable
calculus.
Much of the work has secretly already been done, For example, if xn and yn are real
sequences with limnxn = x and limnyn = y, then in fact limn(xn + yn) = x + y or
limn(xnyn) = xy, because we showed in ?? that arithmetic was continuous. We will also see
that completeness plays a crucial role.
26.1Completeness and inf/sup
Prototypical example for this section: sup[0,1] = sup(0,1) = 1.
As ℝ is a metric space, we may discuss continuity and convergence. There are two important
facts about ℝ which will make most of the following sections tick.
The first fact you have already seen before:
Theorem 26.1.1 (ℝ is complete) As a metric space, ℝ is complete: sequences converge
if and only if they are Cauchy.
The second one we have not seen before — it is the existence of inf and sup. Your intuition
should be:
sup is max adjusted slightly for infinite sets. (And inf is adjusted min.)
Why the “adjustment”?
Example 26.1.2 (Why is max not good enough?) Let’s say we have the open interval
S = (0,1). The elements can get arbitrarily close to 1, so we would like to think “1 is
the max of S”; except the issue is that 1S. In general, infinite sets don’t necessarily
have a maximum, and we have to talk about bounds instead.
So we will define supS in such a way that supS = 1. The definition is that “1 is the
smallest number which is at least every element of S”.
To write it out:
Definition 26.1.3.If S is a set of real numbers:
An upper bound for S is a real number M such that x ≤ M for all x ∈ S. If one
exists, we say S is bounded above;
A lower bound for S is a real number m such that m ≤ x for all x ∈ S. If one
exists, we say S is bounded below.
If both upper and lower bounds exist, we say S is bounded.
Theorem 26.1.4 (ℝ has inf’s and sup’s) Let S be a nonempty set of real numbers.
If S is bounded above then it has a least upper bound, which we denote by supS
and refer to as the supremum of S.
If S is bounded below then it has a greatest lower bound, which we denote by inf S
and refer to as the infimum of S.
Definition 26.1.5.For convenience, if S has not bounded above, we write supS = +∞.
Similarly, if S has not bounded below, we write inf S = −∞.
Example 26.1.6 (Supremums) Since the examples for infimums are basically the same, we
stick with supremums for now.
(a)
If S = then S is not bounded above, so we have supS = +∞.
(b)
If S = denotes the set of negative integers, then supS = −1.
(c)
Let S = [0,1] be a closed interval. Then supS = 1.
(d)
Let S = (0,1) be an open interval. Then supS = 1 as well, even though 1 itself is
not an element of S.
(e)
Let S = ℚ∩(0,1) denote the set of rational numbers between 0 and 1. Then supS = 1
still.
(f)
If S is a finite nonempty set, then supS = maxS.
Definition 26.1.7 (Porting definitions to sequences).If a1, …is a sequence we will often
write
supnan
:= sup
inf nan
: = inf
for the supremum and infimum of the set of elements of the sequence. We also use the words
“bounded above/below” for sequences in the same way.
Example 26.1.8 (Infimum of a sequence) The sequence an = has infimum inf an = 0.
26.2Proofs of the two key completeness properties of ℝ
Careful readers will note that we have not actually proven either ?? or ?? . We will do so
here.
First, we show that the ability to take infimums and supremums lets you prove completeness
of ℝ.
Proof that?? implies?? . Let a1, a2, …be a Cauchy sequence. By discarding finitely
many leading terms, we may as well assume that |ai− aj| ≤ 100 for all i and j. In
particular, the sequence is now bounded; it lies between [a1−100,a1 +100] for example.
We want to show this sequence converges, so we have to first describe what the limit
is. We know that to do this we are really going to have to use the fact that we live in ℝ.
(For example we know in ℚ the limit of 1, 1.4, 1.41, 1.414, …is nonexistent.)
We propose the following: let
We claim that the sequence converges to M = supS.
Exercise 26.2.1.Show that this supremum makes sense by proving that a1− 100 ∈ S (so S isnonempty) while all elements of S are at most a1+ 100 (so S is bounded above). Thus we areallowed to actually take the supremum.
You can think of this set S with the following picture. We have a Cauchy sequence drawn in
the real line which we think converges, which we can visualize as a bunch of dots on the real
line, with some order on them. We wish to cut the line with a knife such that only finitely
many dots are do to the left of the knife. (For example, placing the knife all the way
to the left always works.) The set S represents the places where we could put the
knife, and M is “as far right” as we could go. Because of the way supremums work,
M might not itself be a valid knife location, but certainly anything to its left is.
Let 𝜀 > 0 be given; we want to show eventually all terms are within 𝜀 of M. Because the
sequence is Cauchy, there is an N such that eventually <𝜀 for m ≥ n ≥ N.
Now suppose we fix n and vary m. By the definition of M, it should be possible to
pick the index m such that am≥ M −𝜀 (there are infinitely many to choose from
since M −𝜀 is a valid knife location, and we only need m ≥ n). In that case we
have
by the triangle inequality. This completes the proof. □
Therefore it is enough to prove the latter ?? . To do this though, we would need to
actually give a rigorous definition of the real numbers ℝ, since we have not done so
yet!
One approach that makes this easy to use the so-called Dedekind cut construction.
Suppose we take the rational numbers ℚ. Then one defines a real number to be a “cut” A∣B of
the set of rational numbers: a pair of subsets of ℚ such that
ℚ = A ⊔ B is a disjoint union;
A and B are nonempty;
we have a < b for every a ∈ A and b ∈ B, and
A has no largest element (i.e. supAA).
This can again be visualized by taking what you think of as the real line, and slicing at some real
number. The subset ℚ ⊂ ℝ gets cut into two halves A and B. If the knife happens to land
exactly at a rational number, by convention we consider that number to be in the right half
(which explains the last fourth condition that supAA).
With this definition ?? is easy: to take the supremum of a set of real numbers, we take the
union of all the left halves. The hard part is then figuring out how to define +, −, ×, ÷ and so
on with this rather awkward construction. If you want to read more about this construction in
detail, my favorite reference is [?], in which all of this is done carefully in Chapter
1.
26.3Monotonic sequences
Here is a great exercise.
Exercise 26.3.1 (Mandatory).Prove that if a1≥ a2≥≥ 0 then the limit
exists. Hint: the idea in the proof of the previous section helps; you can also try to usecompleteness of ℝ. Second hint: if you are really stuck, wait until after?? , at which point youcan use essentially copy its proof.
The proof here readily adapts by shifting.
Definition 26.3.2.A sequence an is monotonic if either a1≥ a2≥… or a1≤ a2≤….
Theorem 26.3.3 (Monotonic bounded sequences converge) Let a1, a2, …be a monotonic
bounded sequence. Then limn→∞an exists.
Example 26.3.4 (Silly example of monotonicity) Consider the sequence defined
by
a1
= 1.2
a2
= 1.24
a3
= 1.248
a4
= 1.24816
a5
= 1.2481632
and so on, where in general we stuck on the decimal representation of the next power of 2.
This will converge to some real number, although of course this number is quite unnatural and
there is probably no good description for it.
In general, “infinite decimals” can now be defined as the limit of the truncated finite
ones.
Example 26.3.5 (0.9999 = 1) In particular, I can finally make precise the notion you
argued about in elementary school that
We simply define a repeating decimal to be the limit of the sequence 0.9, 0.99, 0.999….
And it is obvious that the limit of this sequence is 1.
Some of you might be a little surprised since it seems like we really should have
0.9999 = 9 ⋅ 10−1 + 9 ⋅ 10−2 + … — the limit of “partial sums”. Don’t worry, we’re about to
define those in just a moment.
Here is one other great use of monotonic sequences.
Definition 26.3.6.Let a1, a2, …be a sequence (not necessarily monotonic) which is
bounded below. We define
This is called the limit supremum of (an). We set limsupn→∞an to be +∞ if an is
not bounded above.
If an is bounded above, the limit infimum liminf n→∞an is defined similarly. In
particular, liminf n→∞an = −∞ if an is not bounded below.
Exercise 26.3.7.Show that these definitions make sense, by checking that the supremums arenon-increasing, and bounded below.
We can think of limsupnan as “supremum, but allowing finitely many terms to be
discarded”.
26.4Infinite series
Prototypical example for this section:∑k≥1∞ = limn→∞ = 1.
We will actually begin by working with infinite series, since in the previous chapters we
defined limits of sequences, and so this is actually the next closest thing to work
with.1
This will give you a rigorous way to think about statements like
and help answer questions like “how can you add rational numbers and get an irrational
one?”.
Definition 26.4.1.Consider a sequence a1, … of real numbers. The series ∑kakconverges
to a limit L if the sequence of “partial sums”
s1
= a1
s2
= a1 + a2
s3
= a1 + a2 + a3
sn
= a1 + + an
converges to the limit L. Otherwise it diverges.
Abuse of Notation 26.4.2 (Writing divergence as +∞).It is customary, if all the ak
are nonnegative, to write ∑kak = ∞ to denote that the series diverges.
You will notice that by using the definition of sequences, we have masterfully sidestepped
the issue of “adding infinitely many numbers” which would otherwise cause all sorts of
problems.
An “infinite sum” is actually the limit of its partial sums. There is no infiniteaddition involved.
That’s why it’s for example okay to have ∑n≥1 = be irrational; we have already seen
many times that sequences of rational numbers can converge to irrational numbers. It
also means we can gladly ignore all the irritating posts by middle schoolers about
1 + 2 + 3 + = −; the partial sums explode to +∞, end of story, and if you want to assign
a value to that sum it had better be a definition.
Example 26.4.3 (The classical telescoping series) We can now prove the classic
telescoping series
in a way that doesn’t just hand-wave the ending. Note that the kth partial sum
is
∑k=1n
= + + +
= + +
= 1 −.
The limit of this partial sum as n →∞ is 1.
Example 26.4.4 (Harmonic series diverges) We can also make sense of the statement
that ∑k=1∞ = ∞ (i.e. it diverges). We may bound the 2nth partial sums from
below:
∑k=12n
= + + +
≥ + + +
+ + 2n−1 terms
= 1 + + + + = 1 + .
A sequence satisfying s2n≥ 1 + (n − 1) will never converge to a finite number!
I had better also mention that for nonnegative sums, convergence is just the same as having
“finite sum” in the following sense.
Proposition 26.4.5 (Partial sums of nonnegatives bounded implies convergent) Let ∑kak
be a series of nonnegative real numbers. Then ∑kak converges to some limit if and only
if there is a constant M such that
for every positive integer n.
Proof.This is actually just ?? in disguise, but since we left the proof as an exercise
back then, we’ll write it out this time.
Obviously if no such M exists then convergence will not happen, since this means the
sequence sn of partial sums is unbounded.
Conversely, if such M exists then we have s1≤ s2≤< M. Then we contend the
sequence sn converges to L : = supnsn< ∞. (If you read the proof that completeness implies
Cauchy, the picture is nearly the same here, but simpler.)
Indeed, this means for any 𝜀 there are infinitely many terms of the sequence exceeding
L − 𝜀; but since the sequence is monotonic, once sn≥ L − 𝜀 then sn′≥ L − 𝜀 for all n′≥ n.
This implies convergence. □
Abuse of Notation 26.4.6 (Writing ∑< ∞).For this reason, if ak are nonnegative
real numbers, it is customary to write
as a shorthand for “∑kak converges to a finite limit”, (or perhaps shorthand for
“∑kak is bounded” — as we have just proved these are equivalent). We will use this
notation too.
26.5Series addition is not commutative: a horror story
One unfortunate property of the above definition is that it actually depends on the order of
the elements. In fact, it turns out that there is an explicit way to describe when rearrangement
is okay.
Definition 26.5.1.A series ∑kak of real numbers is said to converge absolutely if
i.e. the series of absolute values converges to some limit. If the series converges, but
not absolutely, we say it converges conditionally.
Proposition 26.5.2 (Absolute convergenceconvergence) If a series ∑kak of real
numbers converges absolutely, then it converges in the usual sense.
Exercise 26.5.3 (Great exercise).Prove this by using the Cauchy criteria: show that if thepartial sums of∑k|ak| are Cauchy, then so are the partial sums of∑kak.
Then, rearrangement works great.
Theorem 26.5.4 (Permutation of terms okay for absolute convergence) Consider a series
∑kak which is absolutely convergent and has limit L. Then any permutation of the
terms will also converge to L.
Proof.Suppose ∑kak converges to L, and bn is a rearrangement. Let 𝜀 > 0. We will
show that the partial sums of bn are eventually within 𝜀 of L.
The hypothesis means that there is a large N in terms of 𝜀 such that
for every n ≥ N (the former from vanilla convergence of ak and the latter from the
fact that ak converges absolutely, hence its partial sums are Cauchy).
Now suppose M is large enough that a1, …, aN are contained within the terms {b1,…,bM}.
Then
b1 + + bM
= (a1 + + aN)
+ M − N terms with indices ≥ N
The terms in the first line sum up to within 𝜀 of L, and the terms in the second line have
sum at most 𝜀 in absolute value, so the total b1 + + bM is within 𝜀 + 𝜀 = 𝜀 of
L. □
In particular, when you have nonnegative terms, the world is great:
Nonnegative series can be rearranged at will.
And the good news is that actually, in practice, most of your sums will be nonnegative.
The converse is not true, and in fact, it is almost the worst possible converse you can
imagine.
Theorem 26.5.5 (Permutation of terms meaningless for conditional convergence) Consider a series ∑kak which converges conditionally to some real number. Then, there
exists a permutation of the series which converges conditionally to 1337.
(Or any constant. You can also get it to diverge, too.)
So, permutation is as bad as possible for conditionally convergent series, and hence don’t
even bother to try.
26.6Limits of functions at points
Prototypical example for this section: limx→∞1∕x = 0.
We had also better define the notion of a limit of a real function, which (surprisingly) we
haven’t actually defined yet. The definition will look like what we have seen before with
continuity.
Definition 26.6.1.Let f : ℝ → ℝ be a
function2
and let p ∈ ℝ be a point in the domain. Suppose there exists a real number L such
that:
For every 𝜀 > 0, there exists δ > 0 such that if < δ and x≠p then
< 𝜀.
Then we say L is the limit of f as x → p, and write
There is an important point here: in this definition we deliberately require that x≠p.
The value limx→pf(x) does not depend on f(p), and accordingly we often do noteven bother to define f(p).
Example 26.6.2 (Function with a hole) Define the function f : ℝ → ℝ by
Then limx→0f(x) = 0. The value f(0) = 2019 does not affect the limit. Obviously,
because f(0) was made up to be some artificial value that did not agree with the limit,
this function is discontinuous at x = 0.
Question 26.6.3 (Mandatory).Show that a function f is continuous at p if and only iflimx→pf(x) exists and equals f(p).
Example 26.6.4 (Less trivial example: a rational piecewise function) Define the
function f : ℝ → ℝ as follows:
For example, f(π) = 0, f(2∕3) = , f(0.17) = . Then
For example, if |x| < 1∕100 and x≠0 then f(x) is either zero (for x irrational) or else is
at most (if x is rational).
As f(0) = 1, this function is also discontinuous at x = 0. However, if we change the
definition so that f(0) = 0 instead, then f becomes continuous at 0.
Example 26.6.5 (Famous example) Let f(x) = , f : ℝ → ℝ, where f(0) is assigned
any value. Then
We will not prove this here, since I don’t want to get into trig yet. In general, I will basically
only use trig functions for examples and not for any theory, so most properties of the trig
functions will just be quoted.
Abuse of Notation 26.6.6 (The usual notation). From now on, the above example
will usually be abbreviated to just
The reason there is a slight abuse here is that I’m supposed to feed a function f into
the limit, and instead I’ve written down an expression which is defined everywhere —
except at x = 0. But that f(0) value doesn’t change anything. So the above means: “the
limit of the function described by f(x) = , except f(0) can be whatever it wants
because it doesn’t matter”.
Remark 26.6.7 (For metric spaces) — You might be surprised that I didn’t define the
notion of limx→pf(x) earlier for f : M → N a function on metric spaces. We can actually
do so as above, but there is one nuance: what if our metric space M is discrete, so p has
no points nearby it? (Or even more simply, what if M is a one-point space?) We then
cannot define limx→pf(x) at all.
Thus if f : M → N and we want to define limx→pf(x), we have the requirement that
p should have a point within 𝜀 of it, for any 𝜀 > 0. In other words, p should not be an
isolated point.
As usual, there are no surprises with arithmetic, we have limx→p(f(x)±g(x)) = limx→pf(x)±limx→pg(x),
and so on and so forth. We have effectively done this proof before so we won’t repeat it
again.
26.7Limits of functions at infinity
Annoyingly, we actually have to make this definition separately, even though it will not feel
any different from earlier examples.
Definition 26.7.1.Let f : ℝ → ℝ. Suppose there exists a real number L such that:
For every 𝜀 > 0, there exists a constant M such that if x > M, then
< 𝜀.
Then we say L is the limit of f as x approaches ∞ and write
The limit limx→−∞f(x) is defined similarly, with x > M replaced by x < M.
Fortunately, as ∞ is not an element of ℝ, we don’t have to do the same antics about f(∞) like
we had to do with “f(p) set arbitrarily”. So these examples can be more easily written
down.
Example 26.7.2 (Limit at infinity) The usual:
I’ll even write out the proof: for any 𝜀 > 0, if x > 1∕𝜀 then < 𝜀.
There are no surprises with arithmetic: we have limx→∞(f(x)±g(x)) = limx→∞f(x)±limx→pg(x),
and so on and so forth. This is about the fourth time I’ve mentioned this, so I will not say
more.
26.8A few harder problems to think about
Problem 26A.Define the sequence
for every positive integer n. Compute the limit infimum and the limit supremum.
Problem 26B.For which bounded sequences an does liminf nan = limsupnan?
Problem 26C† (Comparison test).Let ∑an and ∑bn be two series. Assume ∑bn is
absolutely convergent, and |an|≤|bn| for all integers n. Prove that ∑nan is absolutely
convergent.
Problem 26D (Geometric series). Let −1 < r < 1 be a real number. Show that the
series
converges absolutely and determine what it converges to.
Problem 26E (Alternating series test).Let a0≥ a1≥ a2≥ a3≥… be a weakly
decreasing sequence of nonnegative real numbers, and assume that limn→∞an = 0. Show
that the series ∑n(−1)nan is convergent (it need not be absolutely convergent).
Problem 26F ([?, Chapter 3, Exercise 55]).Let (an)n≥1 and (bn)n≥1 be
sequences of real numbers. Assume a1≤ a2≤≤ 1000 and moreover that ∑nbn converges.
Prove that ∑nanbn converges. (Note that in both the hypothesis and statement, we do not
have absolute convergence.)
Problem 26G (Putnam 2016 B1).Let x0,x1,x2,… be the sequence
such that x0 = 1 and for n ≥ 0,
(as usual, log is the natural logarithm). Prove that the infinite series x0 + x1 + … converges
and determine its value.
Problem 26H.Consider again the function f : ℝ → ℝ in ?? defined by
For every real number p, compute limx→pf(x), if it exists. At which points is f continuous?
27Bonus: A hint of p-adic numbers
This is a bonus chapter meant for those who have also read about rings and fields: it’s a nice
tidbit at the intersection of algebra and analysis.
In this chapter, we are going to redo most of the previous chapter with the absolute value
replaced by the p-adic one. This will give us the p-adic integers ℤp, and the p-adic
numbers ℚp. The one-sentence description is that these are “integers/rationals carrying full
mod pe information” (and only that information).
In everything that follows p is always assumed to denote a prime. The first four sections will
cover the founding definitions culminating in a short solution to a USA TST problem. We will
then state (mostly without proof) some more surprising results about continuous functions
f : ℤp→ ℚp; finally we close with the famous proof of the Skolem-Mahler-Lech theorem using
p-adic analysis.
27.1Motivation
Before really telling you what ℤp and ℚp are, let me tell you what you might expect them to
do.
In elementary/olympiad number theory, we’re already well-familiar with the following two
ideas:
Taking modulo a prime p or prime pe, and
Looking at the exponent νp.
Let me expand on the first point. Suppose we have some Diophantine equation. In olympiad
contexts, one can take an equation modulo p to gain something else to work with.
Unfortunately, taking modulo p loses some information: the reduction ℤ ↠ ℤ∕p is far from
injective.
If we want finer control, we could consider instead taking modulo p2, rather than taking
modulo p. This can also give some new information (cubes modulo 9, anyone?), but it has the
disadvantage that ℤ∕p2 isn’t a field, so we lose a lot of the nice algebraic properties that we
got if we take modulo p.
One of the goals of p-adic numbers is that we can get around these two issues I described.
The p-adic numbers we introduce is going to have the following properties:
1.
You can “take modulo pefor all e at once”. In olympiad contexts, we are
used to picking a particular modulus and then seeing what happens if we take
that modulus. But with p-adic numbers, we won’t have to make that choice. An
equation of p-adic numbers carries enough information to take modulo pe.
2.
The numbers ℚpform a field, the nicest possible algebraic structure: 1∕p makes
sense. Contrast this with ℤ∕p2, which is not even an integral domain.
3.
It doesn’t lose as much information as taking modulo p does: rather than the
surjective ℤ ↠ ℤ∕p we have an injective map ℤℤp.
4.
Despite this, you “ignore” some “irrelevant” data. Just like taking modulo
p, you want to zoom-in on a particular type of algebraic information, and this means
necessarily losing sight of other things.1
So, you can think of p-adic numbers as the right tool to use if you only really care about modulo
pe information, but normal ℤ∕pe isn’t quite powerful enough.
To be more concrete, I’ll give a poster example now:
Example 27.1.1 (USA TST 2002/2) For a prime p, show the value of
does not depend on x.
Here is a problem where we clearly only care about pe-type information. Yet it’s a nontrivial
challenge to do the necessary manipulations mod p3 (try it!). The basic issue is that there is
no good way to deal with the denominators modulo p3 (in part ℤ∕p3 is not even an integral
domain).
However, with p-adic analysis we’re going to be able to overcome these limitations and give
a “straightforward” proof by using the identity
Such an identity makes no sense over ℚ or ℝ for convergence reasons, but it will work fine
over the ℚp, which is all we need.
27.2Algebraic perspective
Prototypical example for this section:−1∕2 = 1 + 3 + 32 + 33 + ∈ ℤ3.
We now construct ℤp and ℚp. I promised earlier that a p-adic integer will let you look at
“all residues modulo pe” at once. This definition will formalize this.
27.2.iDefinition of ℤp
Definition 27.2.1 (Introducing ℤp).A p-adic integer is a sequence
of residues xe modulo pe for each integer e, satisfying the compatibility relations
xi≡ xj (mod pi) for i < j.
The set ℤp of p-adic integers forms a ring under component-wise addition and
multiplication.
Example 27.2.2 (Some 3-adic integers) Let p = 3. Every usual integer n generates a
(compatible) sequence of residues modulo pe for each e, so we can view each ordinary
integer as p-adic one:
On the other hand, there are sequences of residues which do not correspond to any usual
integer despite satisfying compatibility relations, such as
which can be thought of as x = 1 + p + p2 + ….
In this way we get an injective map
which is not surjective. So there are more p-adic integers than usual integers.
(Remark for experts: those of you familiar with category theory might recognize that this
definition can be written concisely as
where the inverse limit is taken across e ≥ 1.)
Exercise 27.2.3.Check that ℤpis an integral domain.
27.2.iiBase p expansion
Here is another way to think about p-adic integers using “base p”. As in the example earlier,
every usual integer can be written in base p, for example
More generally, given any x = (x1,…) ∈ ℤp, we can write down a “base p” expansion in the
sense that there are exactly p choices of xk given xk−1. Continuing the example earlier, we
would write
= 1 + 3 + 32 + …
=…11113
and in general we can write
where ak∈{0,…,p − 1}, such that the equation holds modulo pe for each e. Note the
expansion is infinite to the left, which is different from what you’re used to.
(Amusingly, negative integers also have infinite base p expansions: −4 =…2222123,
corresponding to (2 mod 3, 5 mod 9, 23 mod 27, 77 mod 81…).)
Thus you may often hear the advertisement that a p-adic integer is an “possibly infinite
base p expansion”. This is correct, but later on we’ll be thinking of ℤp in a more and more
“analytic” way, and so I prefer to think of this as
p-adic integers are Taylor series with base p.
Indeed, much of your intuition from generating functions K[[X]] (where K is a field) will
carry over to ℤp.
27.2.iiiConstructing ℚp
Here is one way in which your intuition from generating functions carries over:
Proposition 27.2.4 (Non-multiples of p are all invertible) The number x ∈ ℤp is invertible
if and only if x1≠0. In symbols,
Contrast this with the corresponding statement for K[[X]]: a generating function
F ∈ K[[X]] is invertible iff F(0)≠0.
Proof.If x ≡ 0 (mod p) then x1 = 0, so clearly not invertible. Otherwise, xe≢0 (mod p)
for all e, so we can take an inverse ye modulo pe, with xeye≡ 1 (mod pe). As the ye are
themselves compatible, the element (y1,y2,…) is an inverse. □
Example 27.2.5 (We have − =…11113∈ ℤ3) We claim the earlier example is actually
− =
= 1 + 3 + 32 + …
=…11113.
Indeed, multiplying it by −2 gives
(Compare this with the “geometric series” 1 + 3 + 32 + = . We’ll actually be able to
formalize this later, but not yet.)
Remark 27.2.6 (is an integer for p > 2) — The earlier proposition implies that ∈ℤ3 (among other things); your intuition about what is an “integer” is different here!
In olympiad terms, we already knew (mod 3) made sense, which is why calling an
“integer” in the 3-adics is correct, even though it doesn’t correspond to any element of
ℤ.
Exercise 27.2.7 (Unimportant but tricky).Rational numbers correspond exactly to eventuallyperiodic base p expansions.
With this observation, here is now the definition of ℚp.
Definition 27.2.8 (Introducing ℚp).Since ℤp is an integral domain, we let ℚp denote
its field of fractions. These are the p-adic numbers.
Continuing our generating functions analogy:
This means
ℚpcan be thought of as Laurent series with base p.
and in particular according to the earlier proposition we deduce:
Proposition 27.2.9 (ℚplooks like formal Laurent series) Every nonzero element of ℚp is
uniquely of the form
Thus, continuing our base p analogy, elements of ℚp are in bijection with “Laurent
series”
for ak∈. So the base p representations of elements of ℚp can be thought of
as the same as usual, but extending infinitely far to the left (rather than to the
right).
Remark 27.2.10 (Warning) — The field ℚp has characteristic zero, not p.
Remark 27.2.11 (Warning on fraction field) — This result implies that you shouldn’t
think about elements of ℚp as x∕y (for x,y ∈ ℤp) in practice, even though this is the
official definition (and what you’d expect from the name ℚp). The only denominators
you need are powers of p.
To keep pushing the formal Laurent series analogy, K((X)) is usually not thought of
as quotient of generating functions but rather as “formal series with some negative
exponents”. You should apply the same intuition on ℚp.
Remark 27.2.12 — At this point I want to make a remark about the fact 1∕p ∈ ℚp,
connecting it to the wish-list of properties I had before. In elementary number theory
you can take equations modulo p, but if you do the quantity n∕p mod p doesn’t make
sense unless you know n mod p2. You can’t fix this by just taking modulo p2 since then
you need n mod p3 to get n∕p mod p2, ad infinitum. You can work around issues like this,
but the nice feature of ℤp and ℚp is that you have modulo pe information for “all e at
once”: the information of x ∈ ℚp packages all the modulo pe information simultaneously.
So you can divide by p with no repercussions.
27.3Analytic perspective
27.3.iDefinition
Up until now we’ve been thinking about things mostly algebraically, but moving forward it
will be helpful to start using the language of analysis. Usually, two real numbers are
considered “close” if they are close on the number of line, but for p-adic purposes we only care
about modulo pe information. So, we’ll instead think of two elements of ℤp or ℚp as “close” if
they differ by a large multiple of pe.
For this we’ll borrow the familiar νp from elementary number theory.
Definition 27.3.1 (p-adic valuation and absolute value).We define the p-adic valuationνp : ℚp×→ ℤ in the following two equivalent ways:
For x = (x1,x2,…) ∈ ℤp we let νp(x) be the largest e such that xe≡ 0 (mod pe)
(or e = 0 if x ∈ ℤp×). Then extend to all of ℚp× by νp(xy) = νp(x) + νp(y).
Each x ∈ ℚp× can be written uniquely as pku for u ∈ ℤp×, k ∈ ℤ. We let νp(x) = k.
By convention we set νp(0) = +∞. Finally, define the p-adic absolute valuep
by
In particular p = 0.
This fulfills the promise that x and y are close if they look the same modulo pe for large e; in
that case νp(x − y) is large and accordingly p is small.
27.3.iiUltrametric space
In this way, ℚp and ℤp becomes a metric space with metric given by p.
Exercise 27.3.2.Suppose f:ℤp→ ℚpis continuous and f(n) = (−1)nfor every n ∈ ℤ≥0.Prove that p = 2.
In fact, these spaces satisfy a stronger form of the triangle inequality than you are used to
from ℝ.
Proposition 27.3.3 (pis an ultrametric) For any x,y ∈ ℤp, we have the strongtriangle inequality
Equality holds if (but not only if) p≠p.
However, ℚp is more than just a metric space: it is a field, with its own addition and
multiplication. This means we can do analysis just like in ℝ or ℂ: basically, any
notion such as “continuous function”, “convergent series”, et cetera has a p-adic
analog. In particular, we can define what it means for an infinite sum to converge:
Definition 27.3.4 (Convergence notions).Here are some examples of p-adic analogs of
“real-world” notions.
A sequence s1, …converges to a limit L if limn→∞p = 0.
The infinite series ∑kxk converges if the sequence of partial sums s1 = x1,
s2 = x1 + x2, …, converges to some limit.
…et cetera …
With this definition in place, the “base p” discussion we had earlier is now true in the analytic
sense: if x =…a2a1a0p∈ ℤp then
Indeed, the difference between x and the nth partial sum is divisible by pn, hence the partial
sums approach x as n →∞.
While the definitions are all the same, there are some changes in properties that should be
true. For example, in ℚp convergence of partial sums is simpler:
Proposition 27.3.5 (|xk|p→ 0 iff convergence of series) A series ∑k=1∞xk in ℚp
converges to some limit if and only if limk→∞|xk|p = 0.
Contrast this with ∑ = ∞ in ℝ. You can think of this as a consequence of strong triangle
inequality.
Proof.By multiplying by a large enough power of p, we may assume xk∈ ℤp. (This isn’t
actually necessary, but makes the notation nicer.)
Observe that xk (mod p) must eventually stabilize, since for large enough n we have
p< 1 ⇔νp(xn) ≥ 1. So let a1 be the eventual residue modulo p of ∑k=0Nxk
(mod p) for large N. In the same way let a2 be the eventual residue modulo p2, and so
on. Then one can check we approach the limit a = (a1,a2,…). □
27.3.iiiMore fun with geometric series
Let’s finally state the p-adic analog of the geometric series formula.
Proposition 27.3.6 (Geometric series) Let x ∈ ℤp with p< 1. Then
Proof.Note that the partial sums satisfy 1 + x + x2 + + xn = , and xn→ 0 as
n →∞ since p< 1. □
So, 1 + 3 + 32 + = − is really a correct convergence in ℤ3. And so on.
If you buy the analogy that ℤp is generating functions with base p, then all the olympiad
generating functions you might be used to have p-adic analogs. For example, you can prove
more generally that:
Theorem 27.3.7 (Generalized binomial theorem) If x ∈ ℤp and p< 1, then for any
r ∈ ℚ we have the series convergence
(I haven’t defined (1 + x)r, but it has the properties you expect.)
27.3.ivCompleteness
Note that the definition of p could have been given for ℚ as well; we didn’t need ℚp to
introduce it (after all, we have νp in olympiads already). The big important theorem I must
state now is:
Theorem 27.3.8 (ℚpis complete) The space ℚp is the completion of ℚ with respect to
p.
This is the definition of ℚp you’ll see more frequently; one then defines ℤp in terms of ℚp
(rather than vice-versa) according to
27.3.vPhilosophical notes
Let me justify why this definition is philosophically nice. Suppose you are an ancient Greek
mathematician who is given:
Problem for Ancient Greeks. Estimate the value of the sum
to within 0.001.
The sum S consists entirely of rational numbers, so the problem statement would be fair game
for ancient Greece. But it turns out that in order to get a good estimate, it really helps if you
know about the real numbers: because then you can construct the infinite series
∑n≥1n−2 = π2, and deduce that S ≈, up to some small error term from the terms past
, which can be bounded.
Of course, in order to have access to enough theory to prove that S = π2∕6, you need to
have the real numbers; it’s impossible to do calculus in ℚ (the sequence 1, 1.4, 1.41, 1.414, is
considered “not convergent”!)
Now fast-forward to 2002, and suppose you are given
Problem from USA TST 2002. Estimate the sum
to within mod p3.
Even though fp(x) is a rational number, it still helps to be able to do analysis with infinite sums,
and then bound the error term (i.e. take mod p3). But the space ℚ is not complete with
respect to p either, and thus it makes sense to work in the completion of ℚ with respect to
p. This is exactly ℚp.
In any case, let’s finally solve ?? .
Example 27.3.9 (USA TST 2002) We will now compute
Armed with the generalized binomial theorem, this becomes straightforward.
fp(x)
= ∑k=1p−1 = ∑k=1p−1−2
= ∑k=1p−1∑n≥0n
= ∑n≥0∑k=1p−1npn
≡∑k=1p−1− 2xp + 3x2p2 (mod p3).
Using the elementary facts that p2∣∑kk−3 and p∣∑kk−4, this solves the problem.
27.4Mahler coefficients
One of the big surprises of p-adic analysis is that:
We can basically describe all continuous functions ℤp→ ℚp.
They are given by a basis of functions
in the following way.
Theorem 27.4.1 (Mahler; see [?, Theorem 51.1, Exercise 51.b]) Let f : ℤp→ ℚp be continuous,
and define
(27.1)
Then limnan = 0 and
Conversely, if an is any sequence converging to zero, then f(x) = ∑n≥0an defines a
continuous function satisfying (27.1).
The ai are called the Mahler coefficients of f.
Exercise 27.4.2.Last post we proved that if f:ℤp→ℚpis continuous and f(n) = (−1)nforevery n ∈ℤ≥0then p = 2. Re-prove this using Mahler’s theorem, and this time show converselythat a unique such f exists when p = 2.
You’ll note that these are the same finite differences that one uses on polynomials in high
school math contests, which is why they are also called “Mahler differences”.
a0
= f(0)
a1
= f(1) − f(0)
a2
= f(2) − 2f(1) + f(0)
a3
= f(3) − 3f(2) + 3f(1) − f(0).
Thus one can think of an→ 0 as saying that the values of f(0), f(1), …behave like a
polynomial modulo pe for every e ≥ 0.
The notion “analytic” also has a Mahler interpretation. First, the definition.
Definition 27.4.3.We say that a function f : ℤp→ ℚp is analytic if it has a power
series expansion
Theorem 27.4.4 ([?, Theorem 54.4]) The function f(x) = ∑n≥0an is analytic if and
only if
Analytic functions also satisfy the following niceness result:
Theorem 27.4.5 (Strassmann’s theorem) Let f : ℤp→ ℚp be analytic. Then f has finitely
many zeros.
To give an application of these results, we will prove the following result, which was
interesting even before p-adics came along!
Theorem 27.4.6 (Skolem-Mahler-Lech) Let (xi)i≥0 be an integral linear recurrence,
meaning (xi)i≥0 is a sequence of integers
holds for some choice of integers c1, …, ck. Then the set of indices {i∣xi = 0} is eventually
periodic.
Proof.According to the theory of linear recurrences, there exists a matrix A such that
we can write xi as a dot product
Let p be a prime not dividing detA. Let T be an integer such that AT≡ id (mod p)
(with id denoting the identity matrix).
Fix any 0 ≤ r < N. We will prove that either all the terms
are zero, or at most finitely many of them are. This will conclude the proof.
Let AT = id + pB for some integer matrix B. We have
f(n)
= =
= ∑k≥0⋅ pn
= ∑k≥0an where an = pn∈ pnℤ.
Thus we have written f in Mahler form. Initially, we define f : ℤ≥0→ ℤ, but by Mahler’s
theorem (since limnan = 0) it follows that f extends to a function f : ℤp→ ℚp. Also, we can
check that limn = 0 hence f is even analytic.
Thus by Strassman’s theorem, f is either identically zero, or else it has finitely many zeros,
as desired. □
27.5A few harder problems to think about
Problem 27A† (ℤp is compact).Show that ℚp is not compact, but ℤp is. (For the
latter, I recommend using sequential continuity.)
Problem 27B† (Totally disconnected).Show that both ℤp and ℚp are totallydisconnected: there are no connected sets other than the empty set and singleton sets.
Problem 27C (USA TST 2011).Let p be a prime. We say that a sequence of integers
{zn}n=0∞ is a p-pod if for each e ≥ 0, there is an N ≥ 0 such that whenever m ≥ N, pe
divides the sum
Prove that if both sequences {xn}n=0∞ and {yn}n=0∞ are p-pods, then the sequence
{xnyn}n=0∞ is a p-pod.
28Differentiation
28.1Definition
Prototypical example for this section:x3has derivative 3x2.
I suspect most of you have seen this before, but:
Definition 28.1.1.Let U be an open subset1
of ℝ and let f : U → ℝ be a function. Let p ∈ U. We say f is differentiable at p if the
limit2
exists. If so, we denote its value by f′(p) and refer to this as the derivative of f at
p.
The function f is differentiable if it is differentiable at every point. In that case, we
regard the derivative f′: (a,b) → ℝ as a function it its own right.
Exercise 28.1.2.Show that if f is differentiable at p then it is continuous at p too.
Here is the picture. Suppose f : ℝ → ℝ is differentiable (hence continuous). We draw a graph
of f in the usual way and consider values of h. For any nonzero h, what we get is the slope of
the secant line joining (p,f(p)) to (p + h,f(p + h)). However, as h gets close to zero, that
secant line begins to approach a line which is tangent to the graph of the curve. A picture with
f a parabola is shown below, with the tangent in red, and the secant in dashed
green.
So the picture in your head should be that
f′(p) looks like the slope of the tangent line at (p,f(p)).
Remark 28.1.3 — Note that the derivatives are defined for functions on open intervals.
This is important. If f : [a,b] → ℝ for example, we could still define the derivative at
each interior point, but f(a) no longer makes sense since f is not given a value on any
open neighborhood of a.
Let’s do one computation and get on with this.
Example 28.1.4 (Derivative of x3 is 3x2) Let f : ℝ → ℝ by f(x) = x3. For any point p, and
nonzero h we can compute
=
=
= 3p2 + 3ph + h2.
Thus,
Thus the slope at each point of f is given by the formula 3p2. It is customary to then write
f′(x) = 3x2 as the derivative of the entire function f.
Abuse of Notation 28.1.5.We will now be sloppy and write this as (x3)′ = 3x2. This
is shorthand for the significantly more verbose “the real-valued function x3 on domain
so-and-so has derivative 3p2 at every point p in its domain”.
In general, a real-valued differentiable function f : U → ℝ naturally gives rise to
derivative f′(p) at every point p ∈ U, so it is customary to just give up on p altogether
and treat f′ as function itself U → ℝ, even though this real number is of a “different
interpretation”: f′(p) is meant to interpret a slope (e.g. your hourly pay rate) as opposed
to a value (e.g. your total dollar worth at time t). If f is a function from real life, the
units do not even match!
This convention is so deeply entrenched I cannot uproot it without more confusion
than it is worth. But if you read the chapters on multivariable calculus you will see how
it comes back to bite us, when I need to re-define the derivative to be a linear map,
rather than single real numbers.
28.2How to compute them
Same old, right? Sum rule, all that jazz.
Theorem 28.2.1 (Your friendly high school calculus rules) In what follows f and g are
differentiable functions, and U, V are open subsets of ℝ.
(Sum rule) If f,g: U → ℝ then then (f + g)′(x) = f′(x) + g′(x).
(Product rule) If f,g: U → ℝ then then (f ⋅ g)′(x) = f′(x)g(x) + f(x)g′(x).
(Chain rule) If f : U → V and g: V → ℝ then the derivative of the composed
function g ∘ f : U → ℝ is g′(f(x)) ⋅ f′(x).
Proof.
Sum rule: trivial, do it yourself if you care.
Product rule: for every nonzero h and point p ∈ U we may write
which as h → 0 gives the desired expression.
Chain rule: this is where ?? will actually bite us. Let p ∈ U, q = f(p) ∈ V , so
that
We would like to write the expression in the limit as
The problem is that the denominator f(p + h) − f(p) might be zero. So instead,
we define the expression
which is continuous since g was differentiable at q. Then, we do have the equality
because if f(p + h) = q with h≠0, then both sides are equal to zero anyways.
Then, in the limit as h → 0, we have limh→0 = f′(p), while
limh→0Q(f(p + h)) = Q(q) = g′(q) by continuity. This was the desired result. □
Exercise 28.2.2.Compute the derivative of the polynomial f(x) = x3+10x2+2019, viewed asa function f:ℝ →ℝ.
Remark 28.2.3 — Quick linguistic point: the theorems above all hold at each individual
point. For example the sum rule really should say that if f,g: U → ℝ are differentiable
at the point p then so is f + g and the derivative equals f′(p) + g′(p). Thus f and g are
differentiable on all of U, then it of course follows that (f + g)′ = f′ + g′. So each of the
above rules has a “point-by-point” form which then implies the “whole U” form.
We only state the latter since that is what is used in practice. However, in the rare
situations where you have a function differentiable only at certain points of U rather
than the whole interval U, you can still use the below.
We next list some derivatives of well-known functions, but as we do not give rigorous
definitions of these functions, we do not prove these here.
Proposition 28.2.4 (Derivatives of some well-known functions)
The exponential function exp: ℝ → ℝ defined by exp(x) = ex is its own derivative.
The trig functions sin and cos have sin′ = cos, cos′ = −sin.
Example 28.2.5 (A typical high-school calculus question) This means that you can
mechanically compute the derivatives of any artificial function obtained by using the
above, which makes it a great source of busy work in American high schools and
universities. For example, if
then one can compute f′ by:
f′(x)
= (ex)′ + (xsin(x2))′
sum rule
= ex + (xsin(x2))′
above table
= ex + (x)′sin(x2) + x(sin(x2))′
product rule
= ex + sin(x2) + x(sin(x2))′
(x)′ = 1
= ex + sin(x2) + x ⋅ 2x ⋅ cos(x2)
chain rule.
Of course, this function f is totally artificial and has no meaning, which is why calculus is the
topic of widespread scorn in the USA. That said, it is worth appreciating that calculations like
this are possible: it would be better to write the pseudo-theorem “derivatives can actually be
computed”.
If we take for granted that (ex)′ = ex, then we can derive two more useful functions to add
to our library of functions we can differentiate.
Corollary 28.2.6 (Power rule) Let r be a real number. The function ℝ>0→ ℝ by xxr
has derivative (xr)′ = rxr−1.
Proof.We knew this for integers r already, but now we can prove it for any positive real
number r. Write
considered as a function f : ℝ>0→ ℝ. The chain rule (together with the fact that (ex)′ = ex)
now gives
f′(x)
= erlogx⋅ (r log x)′
= erlogx⋅ = xr⋅ = rxr−1.
The reason we don’t prove the formulas for ex and log x is that we don’t at the moment
even have a rigorous definition for either, or even for 2x if x is not rational. However it’s nice
to know that some things imply the other. □
Corollary 28.2.7 (Derivative of log is 1∕x) The function log: ℝ>0→ ℝ has derivative
(log x)′ = 1∕x.
Proof.We have that x = elogx. Differentiate both sides, and again use the chain rule3
Thus (log x)′ = = 1∕x. □
28.3Local (and global) maximums
Prototypical example for this section:Horizontal tangent lines to the parabola are typicallygood pictures.
You may remember from high school that one classical use of calculus was to extract the
minimum or maximum values of functions. We will give a rigorous description of how to do
this here.
Definition 28.3.1.Let f : U → ℝ be a function. A local maximum is a point p ∈ U
such that there exists an open neighborhood V of p (contained inside U) such that
f(p) ≥ f(x) for every x ∈ V .
Definition 28.3.2.A point p is a local extrema if it satisfies either of these.
The nice thing about derivatives is that they pick up all extrema.
Theorem 28.3.3 (Fermat’s theorem on stationary points) Suppose f : U → ℝ is
differentiable and p ∈ U is a local extrema. Then f′(p) = 0.
If you draw a picture, this result is not surprising.
(Note also: the converse is not true. Say, f(x) = x2019 has f′(0) = 0 but x = 0 is not a local
extrema for f.)
Proof.Assume for contradiction f′(p) > 0. Choose any 𝜀 > 0 with 𝜀 < f′(p). Then for
sufficiently small |h| we should have
In particular f(p + h) > f(p) for h > 0 while f(p − h) < f(p) for h < 0. So p is not a
local extremum.
The proof for f′(p) < 0 is similar. □
However, this is not actually adequate if we want a complete method for optimization. The
issue is that we seek global extrema, which may not even exist: for example f(x) = x (which
has f′(x) = 1) obviously has no local extrema at all. The key to resolving this is to use
compactness: we change the domain to be a compact set Z, for which we know that f will
achieve some global maximum. The set Z will naturally have some interior S, and calculus
will give us all the extrema within S. Then we manually check all cases outside
Z.
Let’s see two extended examples. The one is simple, and you probably already know about
it, but I want to show you how to use compactness to argue thoroughly, and how the
“boundary” points naturally show up.
Example 28.3.4 (Rectangle area optimization) Suppose we consider rectangles with
perimeter 20 and want the rectangle with the smallest or largest area.
If we choose the legs of the rectangle to be x and 10 − x, then we are trying to optimize the
function
By compactness, there exists some global maximum and some global minimum.
As f is differentiable on (0,10), we find that for any p ∈ (0,10), a global maximum will be a
local maximum too, and hence should satisfy
Also, the points x = 0 and x = 10 lie in the domain but not the interior (0,10).
Therefore the global extrema (in addition to existing) must be among the three suspects
{0,5,10}.
We finally check f(0) = 0, f(5) = 25, f(10) = 0. So the 5 × 5 square has the largest area and
the degenerate rectangles have the smallest (zero) area.
Here is a non-elementary example.
Proposition 28.3.5 (ex≥ 1 + x) For all real numbers x we have ex≥ 1 + x.
Proof.Define the differentiable function
Consider the compact interval Z = [−1,100]. If x ≤ −1 then obviously f(x) > 0.
Similarly if x ≥ 100 then obviously f(x) > 0 too. So we just want to prove that if x ∈ Z,
we have f(x) ≥ 0.
Indeed, there exists some global minimum p. It could be the endpoints −1 or 100.
Otherwise, if it lies in U = (−1,100) then it would have to satisfy
As f(−1) > 0, f(100) > 0, f(0) = 0, we conclude p = 0 is the global minimum of Z;
and hence f(x) ≥ 0 for all x ∈ Z, hence for all x. □
Remark 28.3.6 — If you are willing to use limits at ±∞, you can rewrite proofs like
the above in such a way that you don’t have to explicitly come up with endpoints like
−1 or 100. We won’t do so here, but it’s nice food for thought.
28.4Rolle and friends
Prototypical example for this section:The racetrack principle, perhaps?
One corollary of the work in the previous section is Rolle’s theorem.
Theorem 28.4.1 (Rolle’s theorem) Suppose f : [a,b] → ℝ is a continuous function, which
is differentiable on the open interval (a,b), such that f(a) = f(b). Then there is a point
c ∈ (a,b) such that f′(c) = 0.
Proof.Assume f is nonconstant (otherwise any c works). By compactness, there exists
both a global maximum and minimum. As f(a) = f(b), either the global maximum or
the global minimum must lie inside the open interval (a,b), and then Fermat’s theorem
on stationary points finishes. □
I was going to draw a picture until I realized xkcd #2042 has one already.
Image from [?]
One can adapt the theorem as follows.
Theorem 28.4.2 (Mean value theorem) Suppose f : [a,b] → ℝ is a continuous function,
which is differentiable on the open interval (a,b). Then there is a point c ∈ (a,b) such
that
Pictorially, there is a c such that the tangent at c has the same slope as the secant joining
(a,f(a)), to (b,f(b)); and Rolle’s theorem is the special case where that secant is
horizontal.
Proof of mean value theorem. Let s = be the slope of the secant line, and
define
which intuitively shears f downwards so that the secant becomes vertical. In fact
g(a) = g(b) now, so we apply Rolle’s theorem to g. □
Remark 28.4.3 (For people with driver’s licenses) — There is a nice real-life
interpretation of this I should mention. A car is travelling along a one-dimensional road
(with f(t) denoting the position at time t). Suppose you cover 900 kilometers in your
car over the course of 5 hours (say f(0) = 0, f(5) = 900). Then there is some point at
time in which your speed at that moment was exactly 180 kilometers per hour, and so
you cannot really complain when the cops pull you over for speeding.
The mean value theorem is important because it lets you relate use derivativeinformation to get information about the function in a way that is really not possible
without it. Here is one quick application to illustrate my point:
Proposition 28.4.4 (Racetrack principle) Let f,g: ℝ → ℝ be two differentiable functions with
f(0) = g(0).
(a)
If f′(x) ≥ g′(x) for every x > 0, then f(x) ≥ g(x) for every x > 0.
(b)
If f′(x) > g′(x) for every x > 0, then f(x) > g(x) for every x > 0.
This proposition might seem obvious. You can think of it as a race track for a reason: if f
and g denote the positions of two cars (or horses etc) and the first car is always faster than the
second car, then the first car should end up ahead of the second car. At a special case g = 0,
this says that if f′(x) ≥ 0, i.e. “f is increasing”, then, well, f(x) ≥ f(0) for x > 0, which had
better be true. However, if you try to prove this by definition from derivatives, you
will find that it is not easy! However, it’s almost a prototype for the mean value
theorem.
Proof of racetrack principle. We prove (a). Let h = f − g, so h(0) = 0. Assume for
contradiction h(p) < 0 for some p > 0. Then the secant joining (0,h(0)) to (p,h(p)) has
negative slope; in other words by mean value theorem there is a 0 < c < p such that
so f′(c) < g′(c), contradiction. Part (b) is the same. □
Sometimes you will be faced with two functions which you cannot easily decouple; the
following form may be more useful in that case.
Theorem 28.4.5 (Ratio mean value theorem) Let f,g: [a,b] → ℝ be two continuous
functions which are differentiable on (a,b), and such that g(a)≠g(b). Then there is a
c ∈ (a,b) such that g′(c)≠0 and
Proof.Use Rolle’s theorem on the function
Remark 28.4.6 — You can capture the case g(a) = g(b) as well if you are willing to
write the conclusion in the less intuitive form g′(c) = f′(c). In
the event g(a) = g(b) then this is just the mean value theorem for g, and the data of f
is irrelevant.
28.5Smooth functions
Prototypical example for this section:All the functions you’re used to.
Let f : U → ℝ be differentiable, thus giving us a function f′: U → ℝ. If our initial function
was nice enough, then we can take the derivative again, giving a function f′′: U → ℝ, and so
on. In general, after taking the derivative n times, we denote the resulting function by f(n). By
convention, f(0) = f.
Definition 28.5.1.A function f : U → ℝ is smooth if it is infinitely differentiable;
that is the function f(n) exists for all n.
Question 28.5.2.Show that the absolute value function is not smooth.
Most of the functions we encounter, such as polynomials, ex, log, sin, cos are smooth,
and so are their compositions. Here is a weird example which we’ll grow more next
time.
Example 28.5.3 (A smooth function with all derivatives zero) Consider the function
This function can be shown to be smooth, with f(n)(0) = 0. So this function has every
derivative at the origin equal to zero, despite being nonconstant!
28.6A few harder problems to think about
Problem 28A (Quotient rule).Let f : (a,b) → ℝ and g: (a,b) → ℝ>0 be differentiable
functions. Let h = f∕g be their quotient (also a function (a,b) → ℝ). Show that the
derivative of h is given by
Problem 28B.For real numbers x > 0, how small can xx be?
Problem 28C (RMM 2018).Determine whether or not there exist
nonconstant polynomials P(x) and Q(x) with real coefficients satisfying
Problem 28D.Let P(x) be a degree n polynomial with real
coefficients. Prove that the equation ex = P(x) has at most n + 1 real solutions in x.
Problem 28E (Jensen’s inequality).Let f : (a,b) → ℝ be a twice differentiable
function such that f′′(x) ≥ 0 for all x (i.e. f is convex). Prove that
for all real numbers x and y in the interval (a,b).
Problem 28F (L’Hôpital rule, or at least one case).Let f,g: ℝ → ℝ be
differentiable functions and let p be a real number. Suppose that
Prove that
provided the right-hand limit exists.
29Power series and Taylor series
Polynomials are very well-behaved functions, and are studied extensively for that reason. From
an analytic perspective, for example, they are smooth, and their derivatives are easy to
compute.
In this chapter we will study power series, which are literally “infinite polynomials”
∑nanxn. Armed with our understanding of series and differentiation, we will see three great
things:
Many of the functions we see in nature actually are given by power series. Among
them are ex, log x, sinx.
Their convergence properties are actually quite well behaved: from the string of
coefficients, we can figure out which x they converge for.
The derivative of ∑nanxn is actually just ∑nnanxn−1.
29.1Motivation
To get the ball rolling, let’s start with one infinite polynomial you’ll recognize: for any fixed
number −1 < x < 1 we have the series convergence
by the geometric series formula.
Let’s pretend we didn’t see this already in ?? . So, we instead have a smooth function
f : (−1,1) → ℝ by
Suppose we wanted to pretend that it was equal to an “infinite polynomial” near the origin,
that is
How could we find that polynomial, if we didn’t already know?
Well, for starters we can first note that by plugging in x = 0 we obviously want
a0 = 1.
We have derivatives, so actually, we can then differentiate both sides to obtain
that
If we now set x = 0, we get a1 = 1. In fact, let’s keep taking derivatives and see what we
get.
(1 − x)−1
=
a0
+
a1x
+
a2x2
+
a3x3
+
a4x4
+
a5x5
+
…
(1 − x)−2
=
a1
+
2a2x
+
3a3x2
+
4a4x3
+
5a5x4
+
…
2(1 − x)−3
=
2a2
+
6a3x
+
12a4x2
+
20a5x3
+
…
6(1 − x)−4
=
6a3
+
24a4x
+
60a5x2
+
…
24(1 − x)−5
=
24a4
+
120a5x
+
…
.
If we set x = 0 we find 1 = a0 = a1 = a2 = … which is what we expect; the geometric series
= 1 + x + x2 + …. And so actually taking derivatives was enough to get the right
claim!
29.2Power series
Prototypical example for this section: = 1 + z + z2 + …, which converges on
(−1,1).
Of course this is not rigorous, since we haven’t described what the right-hand side is, much
less show that it can be differentiated term by term. So we define the main character
now.
Definition 29.2.1.A power series is a sum of the form
where a0, a1, … are real numbers, and z is a variable.
Abuse of Notation 29.2.2 (00 = 1).If you are very careful, you might notice that
when z = 0 and n = 0 we find 00 terms appearing. For this chapter the convention is
that they are all equal to one.
Now, if I plug in a particular real number h, then I get a series of real numbers
∑n=0∞anhn. So I can ask, when does this series converge? It terms out there is a precise
answer for this.
Definition 29.2.3.Given a power series ∑n=0∞anzn, the radius of convergenceR
is defined by the formula
with the convention that R = 0 if the right-hand side is ∞, and R = ∞ if the
right-hand side is zero.
Theorem 29.2.4 (Cauchy-Hadamard theorem) Let ∑n=0∞anzn be a power series with
radius of convergence R. Let h be a real number, and consider the infinite series
of real numbers. Then:
The series converges absolutely if |h| < R.
The series diverges if |h| > R.
Proof.This is not actually hard, but it won’t be essential, so not included. □
Remark 29.2.5 — In the case |h| = R, it could go either way.
Example 29.2.6 (∑zn has radius 1) Consider the geometric series ∑nzn = 1 + z +
z2 + …. Since an = 1 for every n, we get R = 1, which is what we expected.
Therefore, if ∑nanzn is a power series with a nonzero radius R > 0 of convergence, then it
can also be thought of as a function
This is great. Note also that if R = ∞, this means we get a function ℝ → ℝ.
Abuse of Notation 29.2.7 (Power series vs. functions).There is some subtly going
on with “types” of objects again. Analogies with polynomials can help.
Consider P(x) = x3 + 7x + 9, a polynomial. You can, for any real number h, plug in
P(h) to get a real number. However, in the polynomial itself, the symbol x is supposed
to be a variable — which sometimes we will plug in a real number for, but that happens
only after the polynomial is defined.
Despite this, “the polynomial p(x) = x3 + 7x + 9” (which can be thought of as the
coefficients) and “the real-valued function xx3+7x+9” are often used interchangeably.
The same is about to happen with power series: while they were initially thought of
as a sequence of coefficients, the Cauchy-Hadamard theorem lets us think of them as
functions too, and thus we blur the distinction between them.
29.3Differentiating them
Prototypical example for this section:We saw earlier 1 + x + x2 + x3 + …has derivative
1 + 2x + 3x2 + ….
As promised, differentiation works exactly as you want.
Theorem 29.3.1 (Differentiation works term by term) Let ∑n≥0anzn be a power series
with radius of convergence R > 0, and consider the corresponding function
Then all the derivatives of f exist and are given by power series
f′(x)
= ∑n≥1nanxn−1
f′′(x)
= ∑n≥2n(n − 1)anxn−2
which also converge for any x ∈ (−R,R). In particular, f is smooth.
Proof.Also omitted. The right way to prove it is to define the notion “converges
uniformly”, and strengthen Cauchy-Hadamard to have this is as a conclusion as well.
However, we won’t use this later. □
Corollary 29.3.2 (A description of power series coefficients) Let ∑n≥0anzn be a power
series with radius of convergence R > 0, and consider the corresponding function f(x)
as above. Then
Proof.Take the nth derivative and plug in x = 0. □
29.4Analytic functions
Prototypical example for this section:The piecewise e−1∕xor 0 function is not analytic, but issmooth.
With all these nice results about power series, we now have a way to do this process
the other way: suppose that f : U → ℝ is a function. Can we express it as a power
series?
Functions for which this is true are called analytic.
Definition 29.4.1.A function f : U → ℝ is analytic at the point p ∈ U if there exists
an open neighborhood V of p (inside U) and a power series ∑nanzn such that
for any x ∈ V . As usual, the whole function is analytic if it is analytic at each point.
Question 29.4.2.Show that if f is analytic, then it’s smooth.
Moreover, if f is analytic, then by the corollary above its coefficients are actually described
exactly by
Even if f is smooth but not analytic, we can at least write down the power series; we give
this a name.
Definition 29.4.3.For smooth f, the power series ∑n≥0zn is called the Taylorseries of f at p.
Example 29.4.4 (Examples of analytic functions)
(a)
Polynomials, sin, cos, ex, log all turn out to be analytic.
(b)
The smooth function from before defined by
is not analytic. Indeed, suppose for contradiction it was. As all the derivatives are
zero, its Taylor series would be 0 + 0x + 0x2 + …. This Taylor series does converge,
but not to the right value — as f(𝜀) > 0 for any 𝜀 > 0, contradiction.
Theorem 29.4.5 (Analytic iff Taylor series has positive radius) Let f : U → ℝ be a smooth
function. Then f is analytic if and only if for any point p ∈ U, its Taylor series at p has
positive radius of convergence.
Example 29.4.6 It now follows that f(x) = sin(x) is analytic. To see that, we can
compute
f(0)
= sin0 = 0
f′(0)
= cos0 = 1
f′′(0)
= −sin0 = 0
f(3)(0)
= −cos0 = −1
f(4)(0)
= sin0 = 0
f(5)(0)
= cos0 = 1
f(6)(0)
= −sin0 = 0
and so by continuing the pattern (which repeats every four) we find the Taylor series
is
which is seen to have radius of convergence R = ∞.
Like with differentiable functions:
Proposition 29.4.7 (All your usual closure properties for analytic functions) The sums,
products, compositions, nonzero quotients of analytic functions are analytic.
The upshot of this is is that most of your usual functions that occur in nature, or even
artificial ones like f(x) = ex + xsin(x2), will be analytic, hence describable locally by Taylor
series.
29.5A definition of Euler’s constant and exponentiation
We can actually give a definition of ex using the tools we have now.
Definition 29.5.1.We define the map exp: ℝ → ℝ by using the following power series,
which has infinite radius of convergence:
We then define Euler’s constant as e = exp(1).
Question 29.5.2.Show that under this definition,exp′ =exp.
We are then settled with:
Proposition 29.5.3 (exp is multiplicative) Under this definition,
Idea of proof. There is some subtlety here with switching the order of summation that we
won’t address. Modulo that:
exp(x)exp(y)
= ∑n≥0∑m≥0 = ∑n≥0∑m≥0
= ∑k≥0∑ = ∑k≥0∑
= ∑k≥0 = exp(x + y).
□
Corollary 29.5.4 (exp is positive)
(a)
We have exp(x) > 0 for any real number x.
(b)
The function exp is strictly increasing.
Proof.First
which shows exp is nonnegative. Also, 1 = exp(0) = exp(x)exp(−x) implies exp(x)≠0
for any x, proving (a).
(b) is just since exp′ is strictly positive (racetrack principle). □
The log function then comes after.
Definition 29.5.5.We may define log: ℝ>0→ ℝ to be the inverse function of exp.
Since its derivative is 1∕x it is smooth; and then one may compute its coefficients to show it is
analytic.
Note that this actually gives us a rigorous way to define ar for any a > 0 and r > 0,
namely
29.6This all works over complex numbers as well, except also complex analysis is
heaven
We now mention that every theorem we referred to above holds equally well if we work over ℂ,
with essentially no modifications.
Power series are defined by ∑nanzn with an∈ ℂ, rather than an∈ ℝ.
The definition of radius of convergence R is unchanged! The series will converge if
|z| < R.
Differentiation still works great. (The definition of the derivative is unchanged.)
Analytic still works great for functions f : U → ℂ, with U ⊆ ℂ open.
In particular, we can now even define complex exponentials, giving us a function
since the power series still has R = ∞. More generally if a > 0 and z ∈ ℂ we may still
define
(We still require the base a to be a positive real so that log a is defined, though. So this ii
issue is still there.)
However, if one tries to study calculus for complex functions as we did for the real case, in
addition to most results carrying over, we run into a huge surprise:
If f : ℂ → ℂ is differentiable, it is analytic.
And this is just the beginning of the nearly unbelievable results that hold for complex analytic
functions. But this is the part on real analysis, so you will have to read about this
later!
29.7A few harder problems to think about
Problem 29A.Find the Taylor series of log(1 − x).
Problem 29B† (Euler formula).Show that
for any real number 𝜃.
Problem 29C† (Taylor’s theorem, Lagrange form).Let f : [a,b] → ℝ be continuous
and n + 1 times differentiable on (a,b). Define
Prove that there exists ξ ∈ (a,b) such that
This generalizes the mean value theorem (which is the special case n = 0, where P0 = f(a)).
Problem 29D (Putnam 2018 A5).Let f : ℝ → ℝ be smooth, and assume
that f(0) = 0, f(1) = 1, and f(x) ≥ 0 for every real number x. Prove that f(n)(x) < 0 for some
positive integer n and real number x.
30Riemann integrals
“Trying to Riemann integrate discontinuous functions is kind of outdated.” — Dennis Gaitsgory, [?]
We will go ahead and define the Riemann integral, but we won’t do very much with it. The
reason is that the Lebesgue integral is basically better, so we will define it, check the
fundamental theorem of calculus (or rather, leave it as a problem at the end of the chapter),
and then always use Lebesgue integrals forever after.
30.1Uniform continuity
Prototypical example for this section:f(x) = x2is not uniformly continuous on ℝ, butfunctions on compact sets are always uniformly continuous.
Definition 30.1.1.Let f : M → N be a continuous map between two metric spaces.
We say that f is uniformly continuous if for all 𝜀 > 0 there exists a δ > 0 such that
This difference is that given an 𝜀 > 0 we must specify a δ > 0 which works for every choice p
and q of inputs; whereas usually δ is allowed to depend on p and q. (Also, this definition can’t
be ported to a general topological space.)
Example 30.1.2 (Uniform continuity failure)
(a)
The function f : ℝ → ℝ by xx2 is not uniformly continuous. Suppose we take
𝜀 = 0.1 for example. There is no δ such that if |x−y| < δ then |x2−y2| < 0.1, since
as x and y get large, the function f becomes increasingly sensitive to small changes.
(b)
The function (0,1) → ℝ by xx−1 is not uniformly continuous.
(c)
The function ℝ>0→ ℝ by x does turn out to be uniformly continuous (despite
having unbounded derivatives!). Indeed, you can check that the assertion
holds for any x,y,𝜀 > 0.
The good news is that in the compact case all is well.
Theorem 30.1.3 (Uniform continuity free for compact spaces) Let M be a compact metric
space. Then any continuous map f : M → N is also uniformly continuous.
Proof.Assume for contradiction there is some bad 𝜀 > 0. Then taking δ = 1∕n, we find
that for each integer n there exists points pn and qn which are within 1∕n of each other,
but are mapped more than 𝜀 away from each other by f. In symbols, dM(pn,qn) < 1∕n
but dN(f(pn),f(qn)) > 1∕n.
By compactness of M, we can find a convergent subsequence pi1, pi2, … converging to
some x ∈ M.. Since the qin is within 1∕in of pin, it ought to converge as well, to the same
point x ∈ M. Then the sequences f(pin) and f(qin) should both converge to f(x) ∈ N,
but this is impossible as they are always 𝜀 away from each other. □
This means for example that x2 viewed as a continuous function [0,1] → ℝ is automatically
uniformly continuous. Man, isn’t compactness great?
30.2Dense sets and extension
Prototypical example for this section:Functions from ℚ → N extend to ℝ → N if they’reuniformly continuous and N is complete. See also counterexamples below.
Definition 30.2.1.Let S be a subset (or subspace) of a topological space X. Then we
say that S is dense if every open subset of X contains a point of S.
Example 30.2.2 (Dense sets)
(a)
ℚ is dense in ℝ.
(b)
In general, any metric space M is dense in its completion M.
Dense sets lend themselves to having functions completed. The idea is that if I have a
continuous function f : ℚ → N, for some metric space N, then there should be at most one
way to extend it to a function f: ℝ → N. For we can approximate each rational number by
real numbers: if I know f(1), f(1.4), f(1.41), … f() had better be the limit of this sequence.
So it is certainly unique.
However, there are two ways this could go wrong:
Example 30.2.3 (Non-existence of extension)
(a)
It could be that N is not complete, so the limit may not even exist in N. For
example if N = ℚ, then certainly there is no way to extend even the identify function
f : ℚ → N to a function f: ℝ → N.
(b)
Even if N was complete, we might run into issues where f explodes. For example,
let N = ℝ and define
There is also no way to extend this due to the explosion of f near ℚ, which
would cause f() to be undefined.
However, the way to fix this is to require f to be uniformly continuous, and in that case we
do get a unique extension.
Theorem 30.2.4 (Extending uniformly continuous functions) Let M be a metric space, N a
complete metric space, and S a dense subspace of M. Suppose ψ: S → N is a uniformly
continuous function. Then there exists a unique continuous function ψ: M → N such that the
diagram
commutes.
Outline of proof. As mentioned in the discussion, each x ∈ M can be approximated by a
sequence x1, x2, … in S with xi→ x. The two main hypotheses, completeness and uniform
continuity, are now used:
Exercise 30.2.5.Prove that ψ(x1), ψ(x2), … converges in N by using uniform continuity toshow that it is Cauchy, and then appealing to completeness of N.
Hence we define ψ(x) to be the limit of that sequence; this doesn’t depend on the choice of
sequence, and one can use sequential continuity to show ψ is continuous. □
30.3Defining the Riemann integral
Extensions will allow us to define the Riemann integral. I need to introduce a bit of notation
so bear with me.
Definition 30.3.1.Let [a,b] be a closed interval.
We let C0([a,b]) denote the set of continuous functions on [a,b] → ℝ.
We let R([a,b]) denote the set of rectangle functions on [a,b] → ℝ. These
functions which are constant on the intervals [t0,t1), [t1,t2), [t2,t3), …, [tn−2,tn−1),
and also [tn−1,tn], for some a = t0< t1< t2<< tn = b.
We let M([a,b]) = C0([a,b]) ∪ R([a,b]).
Warning: only C0([a,b]) is common notation, and the other two are made up.
See picture below for a typical a rectangle function. (It is irritating that we have to
officially assign a single value to each ti, even though there are naturally two values we
want to use, and so we use the convention of letting the left endpoint be closed).
Definition 30.3.2.We can impose a metric on M([a,b]) by defining
Now, there is a natural notion of integral for rectangle functions: just sum up the obvious
rectangles! Officially, this is the expression
We denote this function by
Theorem 30.3.3 (The Riemann integral) There exists a unique continuous map
such that the diagram
commutes.
Proof.We want to apply the extension theorem, so we just have to check a few things:
We claim R([a,b]) is a dense subset of M([a,b]). In other words, for any continuous
f : [a,b] → ℝ and 𝜀 > 0, we want there to exist a rectangle function that
approximates f within 𝜀.
This follows by uniform continuity. We know there exists a δ > 0 such that
whenever |x−y| < δ we have |f(x) −f(y)| < 𝜀. So as long as we select a rectangle
function whose rectangles have width less than δ, and such that the upper-left
corner of each rectangle lies on the graph of f, then we are all set.
The “add-the-rectangles” map Σ: R([a,b]) → ℝ is uniformly continuous. Actually this
is pretty obvious: if two rectangle functions f and g have d(f,g) < 𝜀, then
d(Σf,Σg) < 𝜀(b − a).
ℝ is complete. □
30.4Meshes
The above definition might seem fantastical, overcomplicated, hilarious, or terrible, depending
on your taste. But if you unravel it, it’s really the picture you are used to. What we have done
is taking every continuous function f : [a,b] → ℝ and showed that it can be approximated by a
rectangle function (which we phrased as a dense inclusion). Then we added the area of the
rectangles. Nonetheless, we will give a definition that’s more like what you’re used to seeing in
other places.
Definition 30.4.1.A tagged partition P of [a,b] consists of a partition of [a,b] into n
intervals, with a point ξi in the nth interval, denoted
The mesh of P is the width of the longest interval, i.e. maxi(ti− ti−1).
Of course the point of this definition is that we add the rectangles, but the ξi are the sample
points.
Theorem 30.4.2 (Riemann integral) Let f : [a,b] → ℝ be continuous. Then
Here the limit means that we can take any sequence of partitions whose mesh approaches
zero.
Proof.The right-hand side corresponds to the areas of some rectangle functions g1,
g2, …with increasingly narrow rectangles. As in the proof ?? , as the meshes of those
rectangles approaches zero, by uniform continuity, we have d(f,gn) → 0 as well. Thus
by continuity in the diagram of ?? , we get limnΣ(gn) = ∫
(f) as needed. □
Combined with the mean value theorem, this can be used to give a short proof of the
fundamental theorem of calculus for functions f with a continuous derivative. The
idea is that for any choice of partition a ≤ t0< t1< t2<< tn≤ b, using the
Mean Value Theorem it should be possible to pick ξi in each interval to match with
the slope of the secant: at which point the areas sum to the total change in f. We
illustrate this situation with three points, and invite the reader to fill in the details as
?? .
One quick note is that although I’ve only defined the Riemann integral for continuous
functions, there ought to be other functions for which it exists (including “piecewise
continuous functions” for example, or functions “continuous almost everywhere”). The
relevant definition is:
Definition 30.4.3.If f : [a,b] → ℝ is a function which is not necessarily continuous,
but for which the limit
exists anyways, then we say f is Riemann integrable on [a,b] and define its value
to be that limit ∫abf(x) dx.
We won’t really use this definition much, because we will see that every Riemann integrable
function is Lebesgue integrable, and the Lebesgue integral is better.
Example 30.4.4 (Your AP calculus returns) We had better mention that ?? implies
that we can compute Riemann integrals in practice, although most of you may already
know this from high-school calculus For example, on the interval (1,4), the derivative
of the function F(x) = x3 is F′(x) = x2. As f(x) = x2 is a continuous function
f : [1,4] → ℝ, we get
Note that we could also have picked F(x) = x3 + 2019; the function F is unique up
to shifting, and this constant cancels out when we subtract. This is why it’s common in
high school to (really) abuse notation and write ∫x2dx = x3 + C.
30.5A few harder problems to think about
Problem 30A.Let f : (a,b) → ℝ be differentiable and assume f′ is bounded. Show that f is
uniformly continuous.
Problem 30B⋆ (Fundamental theorem of calculus). Let f : [a,b] → ℝ be continuous,
differentiable on (a,b), and assume the derivative f′ extends to a continuous function
f′: [a,b] → ℝ. Prove that
Problem 30C⋆ (Improper integrals). For each real number r > 0, evaluate the limit1
or show it does not exist.
This can intuitively be thought of as the “improper” integral ∫01x−rdx; it doesn’t
make sense in our original definition since we did not (and cannot) define the integral
over the non-compact (0,1] but we can still consider the integral over [𝜀,1] for any 𝜀 > 0.
Problem 30D.Show that
Part IX Complex Analysis
31Holomorphic functions
Throughout this chapter, we denote by U an open subset of the complex plane, and by Ω an
open subset which is also simply connected. The main references for this chapter were
[?, ?].
31.1The nicest functions on earth
In high school you were told how to differentiate and integrate real-valued functions. In this
chapter on complex analysis, we’ll extend it to differentiation and integration of
complex-valued functions.
Big deal, you say. Calculus was boring enough. Why do I care about complex
calculus?
Perhaps it’s easiest to motivate things if I compare real analysis to complex analysis. In real
analysis, your input lives inside the real line ℝ. This line is not terribly discerning – you can
construct a lot of unfortunate functions. Here are some examples.
Example 31.1.1 (Optional: evil real functions) You can skim over these very quickly: they’re
only here to make a point.
(a)
The Devil’s Staircase (or Cantor function) is a continuous function H : [0,1] →
[0,1] which has derivative zero “almost everywhere”, yet H(0) = 0 and H(1) = 1.
(b)
The Weierstraß function
is continuous everywhere but differentiable nowhere.
(c)
The function
has the first 99 derivatives but not the 100th one.
(d)
If a function has all derivatives (we call these smooth functions), then it has a
Taylor series. But for real functions that Taylor series might still be wrong. The
function
has derivatives at every point. But if you expand the Taylor series at x = 0, you get
0 + 0x + 0x2 + …, which is wrong for any x > 0 (even x = 0.0001).
Figure 31.1:The Weierstraß Function (image from [?]).
Let’s even put aside the pathology. If I tell you the value of a real smooth function on the
interval [−1,1], that still doesn’t tell you anything about the function as a whole. It
could be literally anything, because it’s somehow possible to “fuse together” smooth
functions.
So what about complex functions? If you consider them as functions ℝ2→ ℝ2, you now
have the interesting property that you can integrate along things that are not line segments:
you can write integrals across curves in the plane. But ℂ has something more: it is a field, so
you can multiply and divide two complex numbers.
So we restrict our attention to differentiable functions called holomorphic functions. It turns
out that the multiplication on ℂ makes all the difference. The primary theme in what follows
is that holomorphic functions are really, really nice, and that knowing tiny amounts of data
about the function can determine all its values.
The two main highlights of this chapter, from which all other results are more or less
corollaries:
Contour integrals of loops are always zero.
A holomorphic function is essentially given by its Taylor series; in particular,
single-differentiable implies infinitely differentiable. Thus, holomorphic functions
behave quite like polynomials.
Some of the resulting corollaries:
It’ll turn out that knowing the values of a holomorphic function on the boundary
of the unit circle will tell you the values in its interior.
Knowing the values of the function at 1, , , …are enough to determine the whole
function!
Bounded holomorphic functions ℂ → ℂ must be constant
And more…
As [?] writes: “Complex analysis is the good twin and real analysis is the evil one: beautiful
formulas and elegant theorems seem to blossom spontaneously in the complex domain, while
toil and pathology rule the reals”.
31.2Complex differentiation
Prototypical example for this section:Polynomials are holomorphic; zis not.
Let f : U → ℂ be a complex function. Then for some z0∈ U, we define the derivative at z0
to be
Note that this limit may not exist; when it does we say f is differentiable at
z0.
What do I mean by a “complex” limit h → 0? It’s what you might expect: for every 𝜀 > 0
there should be a δ > 0 such that
If you like topology, you are encouraged to think of this in terms of open neighborhoods in
the complex plane. (This is why we require U to be open: it makes it possible to take
δ-neighborhoods in it.)
But note that having a complex derivative is actually much stronger than a real function
having a derivative. In the real line, h can only approach zero from below and above, and for
the limit to exist we need the “left limit” to equal the “right limit”. But the complex numbers
form a plane: h can approach zero from many directions, and we need all the limits to be
equal.
Example 31.2.1 (Important: conjugation is not holomorphic) Let f(z) =z be complex
conjugation, f : ℂ → ℂ. This function, despite its simple nature, is not holomorphic!
Indeed, at z = 0 we have,
This does not have a limit as h → 0, because depending on “which direction” we approach
zero from we have different values.
If a function f : U → ℂ is complex differentiable at all the points in its domain it is called
holomorphic. In the special case of a holomorphic function with domain U = ℂ, we call the function
entire.1
Example 31.2.2 (Examples of holomorphic functions) In all the examples below, the
derivative of the function is the same as in their real analogues (e.g. the derivative of ez is
ez).
(a)
Any polynomial zzn + cn−1zn−1 + + c0 is holomorphic.
(b)
The complex exponential exp : x + yiex(cosy + isiny) can be shown to be
holomorphic.
(c)
sin and cos are holomorphic when extended to the complex plane by cosz =
and sinz = .
(d)
As usual, the sum, product, chain rules and so on apply, and hence sums, products,nonzero quotients, and compositions of holomorphic functions are alsoholomorphic.
You are welcome to try and prove these results, but I won’t bother to do so.
31.3Contour integrals
Prototypical example for this section:∮γzmdz around the unit circle.
In the real line we knew how to integrate a function across a line segment [a,b]: essentially,
we’d “follow along” the line segment adding up the values of f we see to get some area. Unlike
in the real line, in the complex plane we have the power to integrate over arbitrary paths: for
example, we might compute an integral around a unit circle. A contour integral lets us
formalize this.
First of all, if f : ℝ → ℂ and f(t) = u(t) + iv(t) for u,v ∈ ℝ, we can define an integral ∫ab
by just adding the real and imaginary parts:
Now let α : [a,b] → ℂ be a path, thought of as a complex
differentiable2
function. Such a path is called a contour, and we define its contour integral
by
You can almost think of this as a u-substitution (which is where the α′ comes from). In
particular, it turns out this integral does not depend on how α is “parametrized”: a circle
given by
and another circle given by
and yet another circle given by
will all give the same contour integral, because the paths they represent have the same
geometric description: “run around the unit circle once”.
In what follows I try to use α for general contours and γ in the special case of
loops.
Let’s see an example of a contour integral.
Theorem 31.3.1 Take γ : [0,2π] → ℂ to be the unit circle specified by
Then for any integer m, we have
Proof.The derivative of eit is ieit. So, by definition the answer is the value of
This is now an elementary calculus question. One can see that this equals 2πi if m = −1 and
otherwise the integrals vanish. □
Let me try to explain why this intuitively ought to be true for m = 0. In that case we have
∮γ1 dz. So as the integral walks around the unit circle, it “sums up” all the tangent vectors at
every point (that’s the direction it’s walking in), multiplied by 1. And given the
nice symmetry of the circle, it should come as no surprise that everything cancels
out. The theorem says that even if we multiply by zm for m≠− 1, we get the same
cancellation.
Definition 31.3.2.Given α : [0,1] → ℂ, we denote by α the “backwards” contour
α(t) = α(1 − t).
Question 31.3.3.What’s the relation between∮αf dz and∮αf dz? Prove it.
This might seem a little boring. Things will get really cool really soon, I promise.
31.4Cauchy-Goursat theorem
Prototypical example for this section:∮γzmdz = 0 for m ≥ 0. But if m < 0, Cauchy’s theoremdoes not apply.
Let Ω ⊆ ℂ be simply connected (for example, Ω = ℂ), and consider two paths α, β with the
same start and end points.
What’s the relation between ∮αf(z) dz and ∮βf(z) dz? You might expect there to be some
relation between them, considering that the space Ω is simply connected. But you probably
wouldn’t expect there to be much of a relation.
As a concrete example, let Ψ : ℂ → ℂ be the function zz − Re[z] (for example,
Ψ(2015 + 3i) = 3i). Let’s consider two paths from −1 to 1. Thus β is walking along the real
axis, and α which follows an upper semicircle.
Obviously ∮βΨ(z) dz = 0. But heaven knows what ∮αΨ(z) dz is supposed to equal.
We can compute it now just out of non-laziness. If you like, you are welcome to
compute it yourself (it’s a little annoying but not hard). If I myself didn’t mess up, it
is
which in particular is not zero.
But somehow Ψ is not a really natural function. It’s not respecting any of the nice,
multiplicative structure of ℂ since it just rudely lops off the real part of its inputs. More
precisely,
Question 31.4.1.Show that Ψ(z) = z−Re[z] is not holomorphic. (Hint: zis not holomorphic.)
Now here’s a miracle: for holomorphic functions, the two integrals are always equal.
Equivalently, (by considering α followed by β) contour integrals of loops are always zero. This
is the celebrated Cauchy-Goursat theorem (also called the Cauchy integral theorem, but later
we’ll have a “Cauchy Integral Formula” so blah).
Theorem 31.4.2 (Cauchy-Goursat theorem) Let γ be a loop, and f : Ω → ℂ a holomorphic
function where Ω is open in ℂ and simply connected. Then
Remark 31.4.3 (Sanity check) — This might look surprising considering that we saw
∮γz−1dz = 2πi earlier. The subtlety is that z−1 is not even defined at z = 0. On the
other hand, the function ℂ ∖{0}→ ℂ by zis holomorphic! The defect now is that
Ω = ℂ ∖{0} is not simply connected. So the theorem passes our sanity checks, albeit
barely.
The typical proof of Cauchy’s Theorem assumes additionally that the partial derivatives of
f are continuous and then applies the so-called Green’s theorem. But it was Goursat who
successfully proved the fully general theorem we’ve stated above, which assumed only
that f was holomorphic. I’ll only outline the proof, and very briefly. You can show
that if f : Ω → ℂ has an antiderivative F : Ω → ℂ which is also holomorphic, and
moreover Ω is simply connected, then you get a “fundamental theorem of calculus”, a
la
where α : [a,b] → ℂ is some path. So to prove Cauchy-Goursat, you only have
to show this antiderivative F exists. Goursat works very hard to prove the result
in the special case that γ is a triangle, and hence by induction for any polygon.
Once he has the result for a rectangle, he uses this special case to construct the
function F explicitly. Goursat then shows that F is holomorphic, completing the
proof.
Anyways, the theorem implies that ∮γzmdz = 0 when m ≥ 0. So much for all our hard
work earlier. But so far we’ve only played with circles. This theorem holds for any contour
which is a loop. So what else can we do?
31.5Cauchy’s integral theorem
We now present a stunning application of Cauchy-Goursat, a “representation theorem”:
essentially, it says that values of f inside a disk are determined by just the values on the
boundary! In fact, we even write down the exact formula. As [?] says, “any time a certain type
of function satisfies some sort of representation theorem, it is likely that many more deep
theorems will follow.” Let’s pull back the curtain:
Theorem 31.5.1 (Cauchy’s integral formula) Let γ : [0,2π] → ℂ be a circle in the plane
given by tReit, which bounds a disk D. Suppose f : U → ℂ is holomorphic such that
U contains the circle and its interior. Then for any point a in the interior of D, we have
Note that we don’t require U to be simply connected, but the reason is pretty silly: we’re
only going to ever integrate f over D, which is an open disk, and hence the disk is simply
connected anyways.
The presence of 2πi, which you saw earlier in the form ∮circlez−1dz, is no accident. In fact,
that’s the central result we’re going to use to prove the result.
Proof.There are several proofs out there, but I want to give the one that really
draws out the power of Cauchy’s theorem. Here’s the picture we have: there’s a
point a sitting inside a circle γ, and we want to get our hands on the value f(a).
We’re going to do a trick: construct a keyhole contour Γδ,𝜀 which has an outer circle γ, plus
an inner circle γ𝜀, which is a circle centered at a with radius 𝜀, running clockwise
(so that γ𝜀 runs counterclockwise). The “width” of the corridor is δ. See picture:
Hence Γδ,𝜀 consists of four smooth curves.
Question 31.5.2.Draw a simply connected open set Ω which contains the entire Γδ,𝜀but doesnot contain the point a.
Hence, the function manages to be holomorphic on all of Ω. Thus Cauchy’s theorem
applies and tells us that
As we let δ → 0, the two walls of the keyhole will cancel each other (because
f is continuous, and the walls run in opposite directions). So taking the limit as
δ → 0, we are left with just γ and γ𝜀, which (taking again orientation into account)
gives
Thus we’ve managed to replace γ with a much smaller circle γ𝜀centered arounda, and the rest is algebra.
To compute the last quantity, write
∮γ𝜀dz
= ∮γ𝜀dz + f(a) ⋅∮γ𝜀dz
= ∮γ𝜀dz + 2πif(a).
where we’ve used ?? Thus, all we have to do is show that
For this we can basically use the weakest bound possible, the so-called ML lemma which I’ll
cite without proof: it says “bound the function everywhere by its maximum”.
Lemma 31.5.3 (ML estimation lemma) Let f be a holomorphic function and α a path.
Suppose M = maxz on α, and let L be the length of α. Then
(This is straightforward to prove if you know the definition of length: L = ∫ab|α′(t)| dt,
where α : [a,b] → ℂ.)
Anyways, as 𝜀 → 0, the quantity approaches f′(a), and so for small enough 𝜀
(i.e. z close to a) there’s some upper bound M. Yet the length of γ𝜀 is the circumference 2π𝜀.
So the ML lemma says that
as desired. □
31.6Holomorphic functions are analytic
Prototypical example for this section:Imagine a formal series∑kckxk!
In the setup of the previous problem, we have a circle γ : [0,2π] → ℂ and a holomorphic
function f : U → ℂ which contains the disk D. We can write
f(a)
= ∮γdz
= ∮γdz
= ∮γf(z)∕z ⋅∑k≥0kdz
You can prove (using the so-called Weierstrass M-test) that the summation order can be
switched:
f(a)
= ∑k≥0∮γ⋅kdz
= ∑k≥0∮γak⋅dz
= ∑k≥0ak.
Letting ck = ∮γdz, and noting this is independent of a, this is
f(a)
= ∑k≥0ckak
and that’s the miracle: holomorphic functions are given by a Taylor series! This is one of
the biggest results in complex analysis. Moreover, if one is willing to believe that we can take
the derivative k times, we obtain
and this gives us f(k)(0) = k! ⋅ ck.
Naturally, we can do this with any circle (not just one centered at zero). So let’s state the
full result below, with arbitrary center p.
Theorem 31.6.1 (Cauchy’s differentiation formula) Let f : U → ℂ be a holomorphic
function and let D be a disk centered at point p bounded by a circle γ. Suppose D is
contained inside U. Then f is given everywhere in D by a Taylor series
where
In particular,
Most importantly,
Over any disk, a holomorphic function is given exactly by a Taylor series.
This establishes a result we stated at the beginning of the chapter: that a function being
complex differentiable once means it is not only infinitely differentiable, but in fact equal to its
Taylor series.
I should maybe emphasize a small subtlety of the result: the Taylor series centered at p is
only valid in a disk centered at p which lies entirely in the domain U. If U = ℂ
this is no issue, since you can make the disk big enough to accommodate any point
you want. It’s more subtle in the case that U is, for example, a square; you can’t
cover the entire square with a disk centered at some point without going outside
the square. However, since U is open we can at any rate at least find some open
neighborhood for which the Taylor series is correct – in stark contrast to the real case.
Indeed, as you’ll see in the problems, the existence of a Taylor series is incredibly
powerful.
31.7A few harder problems to think about
These aren’t olympiad problems, but I think they’re especially nice! In the next complex
analysis chapter we’ll see some more nice applications.
The first few results are the most important.
Problem 31A⋆ (Liouville’s theorem).Let f : ℂ → ℂ be an entire function.
Suppose that < 1000 for all complex numbers z. Prove that f is a constant function.
Problem 31B⋆ (Zeros are isolated).An isolated set in the complex plane is a set of
points S such that around each point in S, one can draw an open neighborhood not
intersecting any other point of S.
Show that the zero set of any nonzero holomorphic function f : U → ℂ is an isolated set,
unless there exists a nonempty open subset of U on which f is identically zero.
Problem 31C⋆ (Identity theorem).Let f,g : U → ℂ be holomorphic, and
assume that U is connected. Prove that if f and g agree on some open neighborhood, then
f = g.
Problem 31D† (Maximums Occur On Boundaries).Let f : U → ℂ be holomorphic,
let Y ⊆ U be compact, and let ∂Y be boundary3
of Y . Show that
In other words, the maximum values of occur on the boundary. (Such maximums
exist by compactness.)
Problem 31E (Harvard quals).Let f : ℂ → ℂ be a nonconstant entire function. Prove
that fimg(ℂ) is dense in ℂ. (In fact, a much stronger result is true: Little Picard’s
theorem says that the image of a nonconstant entire function omits at most one point.)
32Meromorphic functions
32.1The second nicest functions on earth
If holomorphic functions are like polynomials, then meromorphic functions are like rational
functions. Basically, a meromorphic function is a function of the form where
A,B : U → ℂ are holomorphic and B is not zero. The most important example of a
meromorphic function is .
We are going to see that meromorphic functions behave like “almost-holomorphic”
functions. Specifically, a meromorphic function A∕B will be holomorphic at all points except
the zeros of B (called poles). By the identity theorem, there cannot be too many zeros of B!
So meromorphic functions can be thought of as “almost holomorphic” (like , which is
holomorphic everywhere but the origin). We saw that
for γ(t) = eit the unit circle. We will extend our results on contours to such situations.
It turns out that, instead of just getting ∮γf(z) dz = 0 like we did in the holomorphic case,
the contour integrals will actually be used to count the number of poles inside the loop γ. It’s
ridiculous, I know.
32.2Meromorphic functions
Prototypical example for this section:, with a pole of order 1 and residue 1 atz = 0.
Let U be an open subset of ℂ again.
Definition 32.2.1.A function f : U → ℂ is meromorphic if there exists holomorphic
functions A,B: U → ℂ with B not identically zero in any open neighborhood, and
f(z) = A(z)∕B(z) whenever B(z)≠0.
Let’s see how this function f behaves. If z ∈ U has B(z)≠0, then in some small open
neighborhood the function B isn’t zero at all, and thus A∕B is in fact holomorphic; thus f is
holomorphic at z. (Concrete example: is holomorphic in any disk not containing
0.)
On the other hand, suppose p ∈ U has B(p) = 0: without loss of generality, p = 0 to ease
notation. By using the Taylor series at p = 0 we can put
with ck≠0 (certainly some coefficient is nonzero since B is not identically zero!). Then we
can write
But the fraction on the right is a holomorphic function in this open neighborhood! So all
that’s happened is that we have an extra z−k kicking around.
This gives us an equivalent way of viewing meromorphic functions:
Definition 32.2.2.Let f : U → ℂ as usual. A meromorphic function is a function
which is holomorphic on U except at an isolated set S of points (meaning it is
holomorphic as a function U ∖ S → ℂ). For each p ∈ S, called a pole of f, the function
f must admit a Laurent series, meaning that
for all z in some open neighborhood of p, other than z = p. Here m is a positive integer
(and c−m≠0).
Note that the trailing end must terminate. By “isolated set”, I mean that we can draw open
neighborhoods around each pole in S, in such a way that no two open neighborhoods
intersect.
Example 32.2.3 (Example of a meromorphic function) Consider the function
It is meromorphic, because it is holomorphic everywhere except at the zeros of sinz. At each
of these points we can put a Laurent series: for example at z = 0 we have
= (z + 1) ⋅
= ⋅
= ⋅ (z + 1)∑k≥0k.
If we expand out the horrible sum (which I won’t do), then you get times a perfectly fine
Taylor series, i.e. a Laurent series.
Abuse of Notation 32.2.4.We’ll often say something like “consider the function
f : ℂ → ℂ by z”. Of course this isn’t completely correct, because f doesn’t have a
value at z = 0. If I was going to be completely rigorous I would just set f(0) = 2015 or
something and move on with life, but for all intents let’s just think of it as “undefined
at z = 0”.
Why don’t I just write g : ℂ ∖{0}→ ℂ? The reason I have to do this is that it’s still
important for f to remember it’s “trying” to be holomorphic on ℂ, even if isn’t assigned
a value at z = 0. As a function ℂ ∖{0}→ ℂ the function is actually holomorphic.
Remark 32.2.5 — I have shown that any function A(z)∕B(z) has this characterization
with poles, but an important result is that the converse is true too: if f : U ∖ S → ℂ
is holomorphic for some isolated set S, and moreover f admits a Laurent series at each
point in S, then f can be written as a rational quotient of holomorphic functions. I won’t
prove this here, but it is good to be aware of.
Definition 32.2.6.Let p be a pole of a meromorphic function f, with Laurent series
The integer m is called the order of the pole. A pole of order 1 is called a simplepole.
We also give the coefficient c−1 a name, the residue of f at p, which we write Res(f;p).
The order of a pole tells you how “bad” the pole is. The order of a pole is the “opposite”
concept of the multiplicity of a zero. If f has a pole at zero, then its Taylor series near z = 0
might look something like
and so f has a pole of order five. By analogy, if g has a zero at z = 0, it might look
something like
and so g has a zero of multiplicity three. These orders are additive: f(z)g(z) still has a pole
of order 5 − 3 = 2, but f(z)g(z)2 is completely patched now, and in fact has a simple zero
now (that is, a zero of degree 1).
Exercise 32.2.7.Convince yourself that orders are additive as described above. (This is obviousonce you understand that you are multiplying Taylor/Laurent series.)
Metaphorically, poles can be thought of as “negative zeros”.
We can now give many more examples.
Example 32.2.8 (Examples of meromorphic functions)
(a)
Any holomorphic function is a meromorphic function which happens to have no
poles. Stupid, yes.
(b)
The function ℂ → ℂ by z100z−1 for z≠0 but undefined at zero is a meromorphic
function. Its only pole is at zero, which has order 1 and residue 100.
(c)
The function ℂ → ℂ by zz−3 + z2 + z9 is also a meromorphic function. Its only
pole is at zero, and it has order 3, and residue 0.
(d)
The function ℂ → ℂ by z is meromorphic, with the Laurent series at z = 0
given by
Hence the pole z = 0 has order 2 and residue 1.
Example 32.2.9 (A rational meromorphic function) Consider the function ℂ → ℂ given
by
z
= z2 + 1 +
= z2 + 1 + ⋅
= + + (z − 1) + (z − 1)2−…
It has a pole of order 1 and residue 1 at z = 1. (It also has a pole of order 1 at z = −1; you are
invited to compute the residue.)
Example 32.2.10 (Function with infinitely many poles) The function ℂ → ℂ by
has infinitely many poles: the numbers z = 2πk, where k is an integer. Let’s compute the
Laurent series at just z = 0:
=
= ⋅
= ∑k≥0k.
which is a Laurent series, though I have no clue what the coefficients are. You can at least see
the residue; the constant term of that huge sum is 1, so the residue is 1. Also, the pole has
order 1.
The Laurent series, if it exists, is unique (as you might have guessed), and by our result on
holomorphic functions it is actually valid for any disk centered at p (minus the point p). The
part + + is called the principal part, and the rest of the series
c0 + c1(z − p) + … is called the analytic part.
32.3Winding numbers and the residue theorem
Recall that for a counterclockwise circle γ and a point p inside it, we had
where m is an integer. One can extend this result to in fact show that ∮γ(z −p)mdz = 0 for
any loop γ, where m≠− 1. So we associate a special name for the nonzero value at m = −1.
Definition 32.3.1.For a point p ∈ ℂ and a loop γ not passing through it, we define
the winding number, denoted I(p,γ), by
For example, by our previous results we see that if γ is a circle, we have
If you’ve read the chapter on fundamental groups, then this is just the fundamental group
associated to ℂ ∖{p}. In particular, the winding number is always an integer (the proof of this
requires the complex logarithm, so we omit it here). In the simplest case the winding numbers
are either 0 or 1.
Definition 32.3.2.We say a loop γ is regular if I(p,γ) = 1 for all points p in the
interior of γ (for example, if γ is a counterclockwise circle).
With all these ingredients we get a stunning generalization of the Cauchy-Goursat
theorem:
Theorem 32.3.3 (Cauchy’s residue theorem) Let f : Ω → ℂ be meromorphic, where Ω is
simply connected. Then for any loop γ not passing through any of its poles, we have
In particular, if γ is regular then the contour integral is the sum of all the residues, in
the form
Question 32.3.4.Verify that this result coincides with what you expect when you integrate∮γcz−1dz for γ a counter-clockwise circle.
The proof from here is not really too impressive – the “work” was already done in our
statements about the winding number.
Proof.Let the poles with nonzero winding number be p1,…,pk (the others do not affect
the sum).1
Then we can write f in the form
where Pi is the principal part of the pole pi. (For example, if f(z) =
we would write f(z) = (z − 1) + −.)
The point of doing so is that the function g is holomorphic (we’ve removed all the
“bad” parts), so
by Cauchy-Goursat.
On the other hand, if Pi(x) = c1x + c2x2 + + cdxd then
∮γPidz
= ∮γc1⋅dz + ∮γc2⋅2dz + …
= c1⋅I(γ,pi) + 0 + 0 + …
= I(γ,pi)Res(f;pi).
which gives the conclusion. □
32.4Argument principle
One tricky application is as follows. Given a polynomial P(x) = (x−a1)e1(x−a2)e2…(x−an)en,
you might know that we have
The quantity P′∕P is called the logarithmic derivative, as it is the derivative of log P.
This trick allows us to convert zeros of P into poles of P′∕P with order 1; moreover the
residues of these poles are the multiplicities of the roots.
In an analogous fashion, we can obtain a similar result for any meromorphic function
f.
Proposition 32.4.1 (The logarithmic derivative) Let f : U → ℂ be a meromorphic function.
Then the logarithmic derivative f′∕f is meromorphic as a function from U to ℂ; its zeros and
poles are:
(i)
A pole at each zero of f whose residue is the multiplicity, and
(ii)
A pole at each pole of f whose residue is the negative of the pole’s order.
Again, you can almost think of a pole as a zero of negative multiplicity. This spirit is
exemplified below.
Proof.Dead easy with Taylor series. Let a be a zero/pole of f, and WLOG set a = 0
for convenience. We take the Taylor series at zero to get
where k < 0 if 0 is a pole and k > 0 if 0 is a zero. Taking the derivative gives
Now look at f′∕f; with some computation, it equals
So we get a simple pole at z = 0, with residue k. □
Using this trick you can determine the number of zeros and poles inside a regular closed
curve, using the so-called Argument Principle.
Theorem 32.4.2 (Argument principle) Let γ be a regular curve. Suppose f : U → ℂ is
meromorphic inside and on γ, and none of its zeros or poles lie on γ. Then
where Z is the number of zeros inside γ (counted with multiplicity) and P is the number
of poles inside γ (again with multiplicity).
Proof.Immediate by applying Cauchy’s residue theorem alongside the preceding
proposition. In fact you can generalize to any curve γ via the winding number: the
integral is
where the sums are with multiplicity. □
Thus the Argument Principle allows one to count zeros and poles inside any region of
choice.
Computers can use this to get information on functions whose values can be computed but
whose behavior as a whole is hard to understand. Suppose you have a holomorphic function f,
and you want to understand where its zeros are. Then just start picking various circles γ. Even
with machine rounding error, the integral will be close enough to the true integer value that
we can decide how many zeros are in any given circle. Numerical evidence for the Riemann
Hypothesis (concerning the zeros of the Riemann zeta function) can be obtained in this
way.
32.5Philosophy: why are holomorphic functions so nice?
All the fun we’ve had with holomorphic and meromorphic functions comes down to the fact
that complex differentiability is such a strong requirement. It’s a small miracle that
ℂ, which a priori looks only like ℝ2, is in fact a field. Moreover, ℝ2 has the nice
property that one can draw nontrivial loops (it’s also true for real functions that
∫aaf dx = 0, but this is not so interesting!), and this makes the theory much more
interesting.
As another piece of intuition from Siu2 :
If you try to get (left) differentiable functions over quaternions, you find yourself with just
linear functions.
32.6A few harder problems to think about
Problem 32A (Fundamental theorem of algebra).Prove that if f is a nonzero
polynomial of degree n then it has n roots.
Problem 32B† (Rouché’s theorem).Let f,g: U → ℂ be holomorphic functions,
where U contains the unit disk. Suppose that > for all z on the unit circle.
Prove that f and f + g have the same number of zeros which lie strictly inside the unit
circle (counting multiplicities).
Problem 32C (Wedge contour).For each odd integer n, evaluate the
improper integral
Problem 32D (Another contour).Prove that the integral
converges and determine its value.
Problem 32E⋆.Let f : U → ℂ be a nonconstant holomorphic
function.
(a)
(Open mapping theorem) Prove that fimg(U) is open in ℂ.3
(b)
(Maximum modulus principle) Show that cannot have a maximum over U. That
is, show that for any z ∈ U, there is some z′∈ U such that < .
33Holomorphic square roots and logarithms
In this chapter we’ll make sense of a holomorphic square root and logarithm. The main results
are ?? , ?? , ?? , and ?? . If you like, you can read just these four results, and skip the
discussion of how they came to be.
Let f : U → ℂ be a holomorphic function. A holomorphic nth root of f is a function
g : U → ℂ such that f(z) = g(z)n for all z ∈ U. A logarithm of f is a function g : U → ℂ
such that f(z) = eg(z) for all z ∈ U. The main question we’ll try to figure out is: when do these
exist? In particular, what if f = id?
33.1Motivation: square root of a complex number
To start us off, can we define for any complex number z?
The first obvious problem that comes up is that for any z, there are two numbers w such
that w2 = z. How can we pick one to use? For our ordinary square root function, we had a
notion of “positive”, and so we simply took the positive root.
Let’s expand on this: given z = r (here r ≥ 0) we should take the root to
be
such that 2α ≡ 𝜃 (mod 2π); there are two choices for α (mod 2π), differing by
π.
For complex numbers, we don’t have an obvious way to pick α. Nonetheless, perhaps we can
also get away with an arbitrary distinction: let’s see what happens if we just choose the α with
−π < α ≤π.
Pictured below are some points (in red) and their images (in blue) under this “upper-half”
square root. The condition on α means we are forcing the blue points to lie on the right-half
plane.
Here, wi2 = zi for each i, and we are constraining the wi to lie in the right half of the
complex plane. We see there is an obvious issue: there is a big discontinuity near the points w5
and w7! The nearby point w6 has been mapped very far away. This discontinuity occurs since
the points on the negative real axis are at the “boundary”. For example, given −4, we send it
to −2i, but we have hit the boundary: in our interval −π ≤ α <π, we are at the very left
edge.
The negative real axis that we must not touch is what we will later call a branch cut, but for
now I call it a ray of death. It is a warning to the red points: if you cross this line, you will
die! However, if we move the red circle just a little upwards (so that it misses the negative real
axis) this issue is avoided entirely, and we get what seems to be a “nice” square
root.
In fact, the ray of death is fairly arbitrary: it is the set of “boundary issues” that arose when
we picked −π < α ≤π. Suppose we instead insisted on the interval 0 ≤ α < π; then the ray
of death would be the positive real axis instead. The earlier circle we had now works just
fine.
What we see is that picking a particular α-interval leads to a different set of edge cases,
and hence a different ray of death. The only thing these rays have in common is
their starting point of zero. In other words, given a red circle and a restriction of α,
I can make a nice “square rooted” blue circle as long as the ray of death misses
it.
So, what exactly is going on?
33.2Square roots of holomorphic functions
To get a picture of what’s happening, we would like to consider a more general problem: let
f : U → ℂ be holomorphic. Then we want to decide whether there is a g : U → ℂ such
that
Our previous discussion with f = id tells us we cannot hope to achieve this for U = ℂ; there
is a “half-ray” which causes problems. However, there are certainly functions f : ℂ → ℂ such
that a g exists. As a simplest example, f(z) = z2 should definitely have a square
root!
Now let’s see if we can fudge together a square root. Earlier, what we did was try to specify
a rule to force one of the two choices at each point. This is unnecessarily strict. Perhaps we
can do something like: start at a point in z0∈ U, pick a square root w0 of f(z0),
and then try to “fudge” from there the square roots of the other points. What do I
mean by fudge? Well, suppose z1 is a point very close to z0, and we want to pick a
square root w1 of f(z1). While there are two choices, we also would expect w0 to be
close to w1. Unless we are highly unlucky, this should tell us which choice of w1 to
pick. (Stupid concrete example: if I have taken the square root −4.12i of −17 and
then ask you to continue this square root to −16, which sign should you pick for
±4i?)
There are two possible ways we could get unlucky in the scheme above: first, if
w0 = 0, then we’re sunk. But even if we avoid that, we have to worry that if we
run a full loop in the complex plane, we might end up in a different place from
where we started. For concreteness, consider the following situation, again with
f = id:
We started at the point z0, with one of its square roots as w0. We then wound a full red
circle around the origin, only to find that at the end of it, the blue arc is at a different place
where it started!
The interval construction from earlier doesn’t work either: no matter how we pick the
interval for α, any ray of death must hit our red circle. The problem somehow lies with the
fact that we have enclosed the very special point 0.
Nevertheless, we know that if we take f(z) = z2, then we don’t run into any problems with
our “make it up as you go” procedure. So, what exactly is going on?
33.3Covering projections
By now, if you have read the part on algebraic topology, this should all seem quite familiar.
The “fudging” procedure exactly describes the idea of a lifting.
More precisely, recall that there is a covering projection
Let V = . For z ∈ U ∖ V , we already have the square root
g(z) = = = 0. So the burden is completing g : V → ℂ.
Then essentially, what we are trying to do is construct a lifting g in the diagram
Our map p can be described as “winding around twice”. Our ?? now tells us that this
lifting exists if and only if
is a subset of the image of π1(E) by p. Since B and E are both punctured planes, we can
identify them with S1.
Question 33.3.1.Show that the image under p is exactly 2ℤ once we identify π1(B) = ℤ.
That means that for any loop γ in V , we need f ∘γ to have an even winding number around
0 ∈ B. This amounts to
since f has no poles.
Replacing 2 with n and carrying over the discussion gives the first main result.
Theorem 33.3.2 (Existence of holomorphic nth roots) Let f : U → ℂ be holomorphic.
Then f has a holomorphic nth root if and only if
for every contour γ in U.
33.4Complex logarithms
The multivalued nature of the complex logarithm comes from the fact that
So if ew = z, then any complex number w + 2πik is also a solution.
We can handle this in the same way as before: it amounts to a lifting of the following
diagram.
There is no longer a need to work with a separate V since:
Question 33.4.1.Show that if f has any zeros then g can’t possibly exist.
In fact, the map exp : ℂ → ℂ ∖{0} is a universal cover, since ℂ is simply connected. Thus,
pimg(π1(ℂ)) is trivial. So in addition to being zero-free, f cannot have any winding number
around 0 ∈ B at all. In other words:
Theorem 33.4.2 (Existence of logarithms) Let f : U → ℂ be holomorphic. Then f has a
logarithm if and only if
for every contour γ in U.
33.5Some special cases
The most common special case is
Corollary 33.5.1 (Nonvanishing functions from simply connected domains) Let f : Ω → ℂ
be continuous, where Ω is simply connected. If f(z)≠0 for every z ∈ Ω, then f has both
a logarithm and holomorphic nth root.
Finally, let’s return to the question of f = id from the very beginning. What’s the best
domain U such that
is well-defined? Clearly U = ℂ cannot be made to work, but we can do almost as well.
For note that the only zero of f = id is at the origin. Thus if we want to make a
logarithm exist, all we have to do is make an incision in the complex plane that
renders it impossible to make a loop around the origin. The usual choice is to delete
negative half of the real axis, our very first ray of death; we call this a branchcut, with branch point at 0 ∈ ℂ (the point which we cannot circle around). This
gives
Theorem 33.5.2 (Branch cut functions) There exist holomorphic functions
log
: ℂ ∖ (−∞,0] → ℂ
: ℂ ∖ (−∞,0] → ℂ
satisfying the obvious properties.
There are many possible choices of such functions (n choices for the nth root and infinitely
many for log); a choice of such a function is called a branch. So this is what is meant by a
“branch” of a logarithm.
The principal branch is the “canonical” branch, analogous to the way we arbitrarily pick
the positive branch to define : ℝ≥0→ ℝ≥0. For log, we take the w such that ew = z and
the imaginary part of w lies in (−π,π] (since we can shift by integer multiples of 2πi). Often,
authors will write Log z to emphasize this choice.
33.6A few harder problems to think about
Problem 33A.Show that a holomorphic function f : U → ℂ has a holomorphic
logarithm if and only if it has a holomorphic nth root for every integer n.
Problem 33B.Show that the function f : U → ℂ by zz(z − 1) has a holomorphic
square root, where U is the entire complex plane minus the closed interval [0,1].
Part X Measure Theory
34Measure spaces
Here is an outline of where we are going next. Our goal over the next few chapters is to
develop the machinery to state (and in some cases prove) the law of large numbers and the
central limit theorem. For these purposes, the scant amount of work we did in Calculus 101 is
going to be awfully insufficient: integration over ℝ (or even ℝn) is just not going to cut
it.
This chapter will develop the theory of “measure spaces”, which you can think of as “spaces
equipped with a notion of size”. We will then be able to integrate over these with the so-called
Lebesgue integral (which in some senses is almost strictly better than the Riemann
one).
Letter connotations
There are a lot of “types” of objects moving forward, so here are the letter connotations we’ll
use throughout the next several chapters. This makes it easier to tell what the “type” of each
object is just by which letter is used.
Measure spaces denoted by Ω, their elements denoted by ω.
Algebras and σ-algebras denoted by script 𝒜 , ℬ, …. Sets in them denoted by early
capital Roman A, B, C, D, E, ….
Measures (i.e. functions assigning sets to reals) denoted usually by μ or ρ.
Random variables (functions sending worlds to reals) denoted usually by late
capital Roman X, Y , Z, ….
Functions from ℝ → ℝ by Roman letters like f and g for pdf’s and F and G for
cdf’s.
Real numbers denoted by lower Roman letters like x, y, z.
34.1Motivating measure spaces via random variables
To motivate why we want to construct measure spaces, I want to talk about a (real) randomvariable, which you might think of as
the result of a coin flip,
the high temperature in Boston on Saturday,
the possibility of rain on your 18.725 date next weekend.
Why does this need a long theory to develop well? For a simple coin flip one intuitively
just thinks “50% heads, 50% tails” and is done with it. The situation is a little
trickier with temperature since it is continuous rather than discrete, but if all you care
about is that one temperature, calculus seems like it might be enough to deal with
this.
But it gets more slippery once the variables start to “talk to” each other: the high
temperature tells you a little bit about whether it will rain, because e.g. if the temperature is
very high it’s quite likely to be sunny. Suddenly we find ourselves wishing we could talk about
conditional probability, but this is a whole can of worms — the relations between these sorts
of things can get very complicated very quickly.
The big idea to getting a formalism for this is that:
Our measure spaces Ω will be thought of as a space of entire worlds, with eachω ∈ Ω representing a world. Random variables are functions from worlds toℝ.
This way, the space of “worlds” takes care of all the messy interdependence.
Then, we can assign “measures” to sets of worlds: for example, to be a fair coin means that
if you are only interested in that one coin flip, the “fraction” of worlds in which that coin
showed heads should be . This is in some ways backwards from what you were told in
high-school: officially, we start with the space of worlds, rather than starting with the
probabilities.
It will soon be clear that there is no way we can assign a well-defined measure to every
single one of the 2Ω subsets. Fortunately, in practice, we won’t need to, and the
notion of a σ-algebra will capture the idea of “enough measur-able sets for us to get
by”.
Remark 34.1.1 (Random seeds) — Another analogy if you do some programming: each
ω ∈ Ω is a random seed, and everything is determined from there.
34.2Motivating measure spaces geometrically
So, we have a set Ω of possible points (which in the context of the previous discussion can be
thought of as the set of worlds), and we want to assign a measure (think volume) to subsets of
points in Ω. We will now describe some of the obstacles that we will face, in order to motivate
how measure spaces are defined (as the previous section only motivated why we want such
things).
If you try to do this naïvely, you basically immediately run into set-theoretic issues. A
good example to think about why this might happen is if Ω = ℝ2 with the measure
corresponding to area. You can define the area of a triangle as in high school, and you can
then try and define the area of a circle, maybe by approximating it with polygons. But what
area would you assign to the subset ℚ2, for example? (It turns out “zero” is actually a working
answer.) Or, a unit disk is composed of infinitely many points; each of the points
better have measure zero, but why does their union have measure π then? Blah blah
blah.
We’ll say more about this later, but you might have already heard of the Banach-Tarskiparadox which essentially shows there is no good way that you can assign a measure to every
single subset of ℝ3 and still satisfy basic sanity checks. There are just too many possible
subsets of Euclidean space.
However, the good news is that most of these sets are not ones that we will ever care about,
and it’s enough to define measures for certain “sufficiently nice sets”. The adjective we will use
is measurable, and it will turn out that this will be way, way more than good enough for any
practical purposes.
We will generally use A, B, …for measurable sets and denote the entire family of measurable
sets by curly 𝒜 .
34.3σ-algebras and measurable spaces
Here’s the machine code.
Definition 34.3.1.A measurable space consists of a space Ω of points, and a σ-algebra𝒜 of subsets of Ω (the “measurable sets” of Ω). The set 𝒜 is required to satisfy the following
axioms:
𝒜 contains ∅ and Ω.
𝒜 should be closed under complements and countable unions/intersections. (Hint
on nomenclature: σ usually indicates some sort of “countably finite” condition.)
(Complaint: this terminology is phonetically confusing, because it can be confused with
“measure space” later. The way to think about is that “measurable spaces have a σ-algebra, so
we could try to put a measure on it, but we haven’t, yet.”)
Though this definition is how we actually think about it in a few select cases, for
the most part, and we will usually instantiate 𝒜 in practice in a different way:
Definition 34.3.2.Let Ω be a set, and consider some family of subsets ℱ of Ω. Then
the σ-algebra generated by ℱ is the smallest σ-algebra 𝒜 which contains ℱ .
As is commonplace in math, when we see “generated”, this means we sort of let the definition
“take care of itself”. So, if Ω = ℝ, maybe I want 𝒜 to contain all open sets. Well, then the
definition means it should contain all complements too, so it contains all the closed sets. Then
it has to contain all the half-open intervals too, and then…. Rather than try to reason out what
exactly the final shape 𝒜 looks like (which basically turns out to be impossible), we just give
up and say “𝒜 is all the sets you can get if you start with the open sets and apply repeatedly
union/complement operations”. Or even more bluntly: “start with closed sets, shake
vigorously”.
I’ve gone on too long with no examples.
Example 34.3.3 (Examples of measurable spaces) The first two examples actually say what
𝒜 is; the third example (most important) will use generation.
(a)
If Ω is any set, then the power set 𝒜 = 2Ω is obviously a σ-algebra. This will be
used if Ω is countably finite, but it won’t be very helpful if Ω is huge.
(b)
If Ω is an uncountable set, then we can declare 𝒜 to be all subsets of Ω which
are either countable, or which have countable complement. (You should check this
satisfies the definitions.) This is a very “coarse” algebra.
(c)
If Ω is a topological space, the Borel σ-algebra is defined as the σ-algebra generated
by all the open sets of Ω. We denote it by ℬ(Ω), and call the space a Borel space.
As warned earlier, it is basically impossible to describe what it looks like, and instead
you should think of it as saying “we can measure the open sets”.
Question 34.3.4.Show that the closed sets are in ℬ(Ω) for any topological space Ω. Show that[0,1) is also in ℬ(ℝ).
34.4Measure spaces
Definition 34.4.1.Measurable spaces (Ω, 𝒜 ) are then equipped with a function
μ: 𝒜 → [0,+∞] called the measure, which is required to satisfy the following axioms:
μ(∅) = 0
Countable additivity: If A1, A2, …are disjoint sets in 𝒜 , then
The triple (Ω, 𝒜 ,μ) is called a measure space. It’s called a probability space if
μ(Ω) = 1.
Exercise 34.4.2 (Weaker equivalent definitions).I chose to give axioms for 𝒜 and μ that capture howpeople think of them in practice, which means there is some redundancy: for example, being closedunder complements and unions is enough to get intersections, by de Morgan’s law. Here are moreminimal definitions, which are useful if you are trying to prove something satisfies them to reduce theamount of work you have to do:
(a)
The axioms on 𝒜 can be weakened to (i) ∅ ∈𝒜 and (ii) 𝒜 is closed under complementsand countable disjoint unions.
(b)
The axioms on μ can be weakened to (i) μ(∅) = 0, (ii) μ(A ⊔ B) = μ(A) + μ(B), and (iii)for A1⊇ A2⊇, we have μ=limnμ(An).
Remark 34.4.3 — Here are some immediate remarks on these definitions.
If A ⊆ B are measurable, then μ(A) ≤ μ(B) since μ(B) = μ(A) + μ(B − A).
In particular, in a probability space all measures are in [0,1]. On the other hand,
for general measure spaces we’ll allow +∞ as a possible measure (hence the choice
of [0,+∞] as codomain for μ).
We want to allow at least countable unions / additivity because with finite unions
it’s too hard to make progress: it’s too hard to estimate the area of a circle without
being able to talk about limits of countably infinitely many triangles.
We don’t want to allow uncountable unions and additivity, because uncountable
sums basically never work out. In particular, there is a nice elementary exercise as
follows:
Exercise 34.4.4 (Tricky).Let S be an uncountable set of positive real numbers. Show thatsome finite subset T ⊆ S has sum greater than 102019. Colloquially, “uncountable many positivereals cannot have finite sum”.
So countable sums are as far as we’ll let the infinite sums go. This is the reason why we
considered σ-algebras in the first place.
Example 34.4.5 (Measures) We now discuss measures on each of the spaces in our previous
examples.
(a)
If 𝒜 = 2Ω (or for that matter any 𝒜 ) we may declare μ(A) = |A| for each A ∈ 𝒜
(even if |A| = ∞). This is called the counting measure, simply counting the
number of elements.
This is useful if Ω is countably infinite, and optimal if Ω is finite (and nonempty). In
the latter case, we will often normalize by μ(A) = so that Ω becomes a probability
space.
(b)
Suppose Ω was uncountable and we took 𝒜 to be the countable sets and their
complements. Then
is a measure. (Check this.)
(c)
Elephant in the room: defining a measure on ℬ(Ω) is hard even for Ω = ℝ, and is
done in the next chapter. So you will have to hold your breath. Right now, all you
know is that by declaring my intent to define a measure ℬ(Ω), I am hoping that at
least every open set will have a volume.
34.5A hint of Banach-Tarski
I will now try to convince you that ℬ(Ω) is a necessary concession, and for general topological
spaces like Ω = ℝn, there is no hope of assigning a measure to 2Ω. (In the literature, this
example is called a Vitali set.)
Example 34.5.1 (A geometric example why 𝒜 = 2Ω is unsuitable) Let Ω denote the
unit circle in ℝ2 and 𝒜 = 2Ω. We will show that any measure μ on Ω with μ(Ω) = 1
will have undesirable properties.
Let ∼ denote an equivalence relation on Ω defined as follows: two points are equivalent
if they differ by a rotation around the origin by a rational multiple of π. We may pick a
representative from each equivalence class, letting X denote the set of representatives.
Then
Since we’ve only rotated X, each of the rotations should have the same measure m. But
μ(Ω) = 1, and there is no value we can assign that measure: if m = 0 we get μ(Ω) = 0
and m > 0 we get μ(Ω) = ∞.
Remark 34.5.2 (Choice) — Experts may recognize that picking a representative
(i.e. creating set X) technically requires the Axiom of Choice. That is why, when people
talk about Banach-Tarski issues, the Axiom of Choice almost always gets honorable
mention as well.
Stay tuned to actually see a construction for ℬ(ℝn) in the next chapter.
34.6Measurable functions
In the past, when we had topological spaces, we considered continuous functions. The analog
here is:
Definition 34.6.1.Let (X, 𝒜 ) and (Y, ℬ) be measurable spaces (or measure spaces).
A function f : X → Y is measurable if for any measurable set S ⊆ Y (i.e. S ∈ ℬ) we
have fpre(S) is measurable (i.e. fpre(S) ∈ 𝒜 ).
In practice, most functions you encounter will be continuous anyways, and in that case we
are fine.
Proposition 34.6.2 (Continuous implies Borel measurable) Suppose X and Y
are topological spaces and we pick the Borel measures on both. A function f : X → Y
which is continuous as a map of topological spaces is also measurable.
Proof.Follows from the fact that pre-images of open sets are open, and the Borel measure
is generated by open sets. □
34.7On the word “almost” (TO DO)
In later chapters we will begin seeing the phrase “almost everywhere” and “almost
surely” start to come up, and it seems prudent to take the time to talk about it
now.
Definition 34.7.1.We say that property P occurs almost everywhere or almostsurely if the set
has measure zero.
For example, if we say “f = g almost everywhere” for some functions f and g defined on a
measure space Ω, then we mean that f(ω) = g(ω) for all ω ∈ Ω other than a measure-zero
set.
There, that’s the definition. The main thing to now update your instincts on is
that
In measure theory, we basically only care about things up to almost-everywhere.
Here are some examples:
If f = g almost everywhere, then measure theory will basically not tell these
functions apart. For example, ∫Ωf dω = ∫Ωg dω will hold for two functions
agreeing almost everywhere.
As another example, if we prove “there exists a unique function f such that
so-and-so”, the uniqueness is usually going to be up to measure-zero sets.
You can think of this sort of like group isomorphism, where two groups are considered “basically
the same” when they are isomorphic, except this one might take a little while to get used
to.1
34.8A few harder problems to think about
Problem 34A†.Let (Ω, 𝒜 ,μ) be a probability space. Show that the intersection of
countably many sets of measure 1 also has measure 1.
Problem 34B (On countable σ-algebras).Let 𝒜 be a σ-algebra on
a set Ω. Suppose that 𝒜 has countable cardinality. Prove that |𝒜 | is finite and equals
a power of 2.
35Constructing the Borel and Lebesgue measure
It’s very difficult to define in one breath a measure on the Borel space ℬ(ℝn). It is easier if we
define a weaker notion first. There are two such weaker notions that we will define:
A pre-measure: satisfies the axioms of a measure, but defined on fewer sets
than a measure: they’ll be defined on an “algebra” rather than the full-fledged
“σ-algebra”.
An outer measure: defined on 2Ω but satisfies weaker axioms.
It will turn out that pre-measures yield outer measures, and outer measures yield
measures.
35.1Pre-measures
Prototypical example for this section:Let Ω = ℝ2. Then we take 𝒜0generated by rectangles,with μ0the usual area.
The way to define a pre-measure is to weaken the σ-algebra to an algebra.
Definition 35.1.1.Let Ω be a set. We define notions of an algebra, which is the same
as σ-algebra except with “countable” replaced by finite everywhere.
That is: an algebra 𝒜0 on Ω is a nonempty subset of 2Ω, which is closed under
complement and finite union. The smallest algebra containing a subset ℱ ⊆ 2Ω is the
algebra generated by ℱ .
In practice, we will basically always use generation for algebras.
Example 35.1.2 When Ω = ℝn, we can let ℒ0 be the algebra generated by [a1,b1] ×× [an,bn].
A typical element might look like:
Unsurprisingly, since we have finitely many rectangles and their complements involved, in this
case we actually can unambiguously assign an area, and will do so soon.
Definition 35.1.3.A pre-measureμ0 on a algebra 𝒜0 is a function μ0: 𝒜0→ [0,+∞]
which satisfies the axioms
μ0(∅) = 0, and
Countable additivity: if A1, A2, …are disjoint sets in 𝒜0 and moreover the
disjoint union ⊔Ai is contained in 𝒜0 (not guaranteed by algebra axioms!), then
Example 35.1.4 (The pre-measure on ℝn) Let Ω = ℝ2. Then, let ℒ0 be the algebra
generated by rectangles [a1,a2] × [b1,b2]. We then let
the area of the rectangle. As elements of ℒ0 are simply finite unions of rectangles and their
complements (picture drawn earlier), it’s not difficult to extend this to a pre-measure λ0
which behaves as you expect — although we won’t do this.
Since we are sweeping something under the rug that turns out to be conceptually important,
I’ll go ahead and blue-box it.
Proposition 35.1.5 (Geometry sanity check that we won’t prove) For Ω = ℝn and ℒ0 the
algebra generated by rectangular prisms, one can define a pre-measure λ0 on ℒ0.
From this point forwards, we will basically do almost no
geometry1
whatsoever in defining the measure ℬ(ℝn), and only use set theory to extend our measure. So,
?? is the only sentry which checks to make sure that our “initial definition” is
sane.
To put the point another way, suppose an insanescientist2
tried to define a notion of area in which every rectangle had area 1. Intuitively, this shouldn’t
be possible: every rectangle can be dissected into two halves and we ought to have 1 + 1≠1.
However, the only thing that would stop them is that they couldn’t extend their
pre-measure on the algebra ℒ0. If they somehow got past that barrier and got a
pre-measure, nothing in the rest of the section would prevent them from getting
an entire bona fide measure with this property. Thus, in our construction of the
Lebesgue measure, most of the geometric work is captured in the (omitted) proof of
?? .
35.2Outer measures
Prototypical example for this section:Keep taking Ω = ℝ2; see the picture to follow.
The other way to weaken a measure is to relax the countable additivity, and this yields the
following:
Definition 35.2.1.An outer measureμ∗ on a set Ω is a function μ∗: 2Ω→ [0,+∞]
satisfying the following axioms:
μ∗(∅) = 0;
if E ⊆ F and E,F ∈ 2Ω then μ∗(E) ≤ μ∗(F);
for any subsets E1, E2, …of Ω we have
(I don’t really like the word “outer measure”, since I think it is a bit of a misnomer: I would
rather call it “fake measure”, since it’s not a measure either.)
The reason for the name “outer measure” is that you almost always obtain outer measures
by approximating them from “outside” sets. Officially, the result is often stated as follows (as
?? ).
For a set Ω, let ℰ be any subset of 2Ω and let ρ: ℰ→ [0,+∞] be any function.
Then
is an outer measure.
However, I think the above theorem is basically always wrong to use in practice, because it
is way too general. As I warned with the insane scientist, we really do want some sort of sanity
conditions on ρ: otherwise, if we apply the above result as stated, there is no guarantee that μ∗
will be compatible with ρ in any way.
So, I think it is really better to apply the theorem to pre-measures μ0 for which one does
have some sort of guarantee that the resulting μ∗ is compatible with μ0. In practice, this is
always how we will want to construct our outer measures.
Theorem 35.2.2 (Constructing outer measures from pre-measures) Let μ0 be a pre-measure on
an algebra 𝒜0 on a set Ω.
(a)
The map μ∗: 2Ω→ [0,+∞] defined by
is an outer measure.
(b)
Moreover, this measure agrees with μ0 on sets in 𝒜0.
Intuitively, what is going on is that μ∗(A) is the infimum of coverings of A by countable
unions of elements in 𝒜0. Part (b) is the first half of the compatibility condition I promised;
the other half appears later as ?? .
Proof of?? . As alluded to already, part (a) is a special case of ?? (and proving it
in this generality is actually easier, because you won’t be distracted by unnecessary
properties).
We now check (b), that μ∗(A) = μ0(A) for A ∈ 𝒜0. One bound is quick:
Question 35.2.3.Show that μ∗(A) ≤ μ0(A).
For the reverse, suppose that A ⊆⋃nAn. Then, define the sets
B1
= A ∩ A1
B2
= (A ∩ A2) ∖ B1
B3
= (A ∩ A3) ∖ B2
and so on. Then the Bn are disjoint elements of 𝒜0 with Bn⊂ An, and we have rigged the
definition so that ⊔nBn = A. Thus by definition of pre-measure,
as desired. □
Example 35.2.4 Let Ω = ℝ2 and λ0 the pre-measure from before. Then λ∗(A) is,
intuitively, the infimum of coverings of the set A by rectangles. Here is a picture you
might use to imagine the situation with A being the unit disk.
35.3Carathéodory extension for outer measures
We will now take any outer measure and turn it into a proper measure. To do this, we first
need to specify the σ-algebra on which we will define the measure.
Definition 35.3.1.Let μ∗ be an outer measure. We say a set A is Carathéodorymeasurable with respect to μ∗, or just μ∗-measurable, if the following condition
holds: for any set E ∈ 2Ω,
This definition is hard to motivate, but turns out to be the right one. One way to
motivate is this: it turns out that in ℝn, it will be equivalent to a reasonable geometric
condition (which I will state in ?? ), but since that geometric definition requires
information about ℝn itself, this is the “right” generalization for general measure
spaces.
Since our goal was to extend our 𝒜0, we had better make sure this definition lets us
measure the initial sets that we started with!
Proposition 35.3.2 (Carathéodory measurability is compatible with the initial 𝒜0) Suppose μ∗ was obtained from a pre-measure μ0 on an algebra 𝒜0, as in ?? . Then
every set in 𝒜0 is μ∗-measurable.
This is the second half of the compatibility condition that we get if we make sure
our initial μ0 at least satisfies the pre-measure axioms. (The first half was (b) of
?? .)
Proof.Let A ∈ 𝒜0 and E ∈ 2Ω; we wish to prove μ∗(E) = μ∗(E ∩ A) + μ∗(E ∖ A). The
definition of outer measure already requires μ∗(E) ≤ μ∗(E ∩ A) + μ∗(E ∖ A) and so it’s
enough to prove the reverse inequality.
By definition of infimum, for any 𝜀 > 0, there is a covering E ⊂⋃nAn with
μ∗(E) + 𝜀 ≥∑nμ0(An). But
with the first equality being the definition of pre-measure on 𝒜0, the second just
being by definition of μ∗ (since An∩ A certainly covers E ∩ A, for example). Thus
μ∗(E) + 𝜀 ≥ μ∗(E ∩ A) + μ∗(E ∖ A). Since the inequality holds for any 𝜀 > 0, we’re
done. □
To add extra icing onto the cake, here is one more niceness condition which our constructed
measure will happen to satisfy.
Definition 35.3.3.A null set of a measure space (Ω, 𝒜 ,μ) is a set A ∈ 𝒜 with
μ(A) = 0. A measure space (Ω, 𝒜 ,μ) is complete if whenever A is a null set, then all
subsets of A are in 𝒜 as well (and hence null sets).
This is a nice property to have, for obvious reasons. Visually, if I have a bunch of dust which I
already assigned weight zero, and I blow away some of the dust, then the remainder should
still have an assigned weight — zero. The extension theorem will give us σ-algebras with this
property.
Theorem 35.3.4 (Carathéodory extension theorem for outer measures) If μ∗ is an outer
measure, and 𝒜 cm is the set of μ∗-measurable sets with respect to μ∗, then 𝒜 cm is a
σ-algebra on Ω, and the restriction μcm of μ∗ to 𝒜 cm gives a complete measure space.
(Phonetic remark: you can think of the superscript cm as standing for either
“Carathéodory measurable” or “complete”. Both are helpful for remembering what this
represents. This notation is not standard but the pun was too good to resist.)
Thus, if we compose ?? with ?? , we find that every pre-measure μ0 on an algebra 𝒜0
naturally gives a σ-algebra 𝒜 cm with a complete measure μcm, and our two compatibility
results (namely (b) of ?? , together with ?? ) means that 𝒜 cm⊃𝒜0 and μcm agrees with
μ.
Here is a table showing the process, where going down each row of the table corresponds to
restriction process.
Construct order
Notes
2Ω
μ∗
Step 2
μ∗ is outer measure obtained from μ0
𝒜 cm
μcm
Step 3
𝒜 cm defined as μ∗-measurable sets,
(𝒜 cm,μcm)
is complete.
𝒜0
μ0
Step 1
μ0 is a pre-measure
35.4Defining the Lebesgue measure
This lets us finally define the Lebesgue measure on ℝn. We wrap everything together at once
now.
Definition 35.4.1.We create a measure on ℝn by the following procedure.
Start with the algebra ℒ0 generated by rectangular prisms, and define a
pre-measure λ0 on this ℒ0 (this was glossed over in the example).
By ?? , this gives the Lebesgue outer measureλ∗ on 2ℝn, which is compatible
on all the rectangular prisms.
By Carathéodory (?? ), this restricts to a complete measure λ on the σ-algebra
ℒ(ℝn) of λ∗-measurable sets (which as promised contains all rectangular prisms).3
The resulting complete measure, denoted λ, is called the Lebesgue measure.
The algebra ℒ(ℝn) we obtained will be called the Lebesgue σ-algebra; sets in it are said
to be Lebesgue measurable.
Here is the same table from before, with the values filled in for the special case Ω = ℝn,
which gives us the Lebesgue algebra.
Construct order
Notes
2ℝn
λ∗
Step 2
λ∗ is Lebesgue outer measure
ℒ(ℝn)
λ
Step 3
Lebesgue σ-algebra (complete)
ℒ0
λ0
Step 1
Define pre-measure on rectangles
Of course, now that we’ve gotten all the way here, if we actually want to compute any
measures, we can mostly gleefully forget about how we actually constructed the measure and
just use the properties. The hard part was to showing that there is a way to assign measures
consistently; actually figuring out what that measure’s value is given that it exists is often
much easier. Here is an example.
Example 35.4.2 (The Cantor set has measure zero) The standard middle-thirdsCantor set is the subset [0,1] obtained as follows: we first delete the open interval
(1∕3,2∕3). This leaves two intervals [0,1∕3] and [2∕3,1] from which we delete the
middle thirds again from both, i.e. deleting (1∕9,2∕9) and (7∕9,8∕9). We repeat this
procedure indefinitely and let C denote the result. An illustration is shown below.
Image from [?]
It is a classic fact that C is uncountable (it consists of ternary expansions omitting the digit
1). But it is measurable (it is an intersection of closed sets!) and we contend it has measure
zero. Indeed, at the nth step, the result has measure (2∕3)n leftover. So μ(C) ≤ (2∕3)n for
every n, forcing μ(C) = 0.
This is fantastic, but there is one elephant in the room: how are the Lebesgue σ-algebra and
the Borel σ-algebra related? To answer this question briefly, I will state two results (but
another answer is given in the next section). The first is a geometric interpretation of the
strange Carathéodory measurable hypothesis.
Proposition 35.4.3 (A geometric interpretation of Lebesgue measurability) A set A ⊆ ℝn
is Lebesgue measurable if and only if for every 𝜀 > 0, there is an open set U ⊃ A such
that
where λ∗ is the Lebesgue outer measure.
I want to say that this was Lebesgue’s original formulation of “measurable”, but I’m not
sure about that. In any case, we won’t need to use this, but it’s good to see that our definition
of Lebesgue measurable has a down-to-earth geometric interpretation.
Question 35.4.4.Deduce that every open set is Lebesgue measurable. Conclude that theLebesgue σ-algebra contains the Borel σ-algebra. (A different proof is given later on.)
However, the containment is proper: there are more Lebesgue measurable sets than Borel
ones. Indeed, it can actually be proven using transfinite induction (though we won’t) that
= . Using this, one obtains:
Exercise 35.4.5.Show the Borel σ-algebra is not complete. (Hint: consider the Cantor set. Youwon’t be able to write down an example of a non-measurable set, but you can use cardinalityarguments.) Thus the Lebesgue σ-algebra strictly contains the Borel one.
Nonetheless, there is a great way to describe the Lebesgue σ-algebra, using the idea of
completeness.
Definition 35.4.6.Let (Ω, 𝒜 ,μ) be a measure space. The completion (Ω,𝒜,μ) is
defined as follows: we let
and μ(A ∪ N) = μ(A). One can check this is well-defined, and in fact μ is the unique
extension of μ from 𝒜 to 𝒜.
This looks more complicated than it is. Intuitively, all we are doing is “completing”
the measure by telling μ to regard any subset of a null set as having measure zero, too.
Then, the saving grace:
Theorem 35.4.7 (Lebesgue is completion of Borel) For ℝn, the Lebesgue measure is the
completion of the Borel measure.
Proof.This actually follows from results in the next section, namely ?? and part (c) of
Carathéodory for pre-measures (?? ). □
35.5A fourth row: Carathéodory for pre-measures
Prototypical example for this section:The fourth row for the Lebesgue measure isℬ(ℝn).
In many cases, 𝒜 cm is actually bigger than our original goal, and instead we only need to
extend μ0 on 𝒜0 to μ on 𝒜 , where 𝒜 is the σ-algebra generated by 𝒜0. Indeed, our original
goal was to get ℬ(ℝn), and in fact:
Exercise 35.5.1.Show that ℬ(ℝn) is the σ-algebra generated by the ℒ0we defined earlier.
Fortunately, this restriction is trivial to do.
Question 35.5.2.Show that 𝒜cm⊃𝒜, so we can just restrict μcmto 𝒜 .
We will in a moment add this as the fourth row in our table.
However, if this is the end goal, than a somewhat different Carathéodory theorem can be
stated because often one more niceness condition holds:
Definition 35.5.3.A pre-measure or measure μ on Ω is σ-finite if Ω can be written
as a countable union Ω = ⋃nAn with μ(An) < ∞ for each n.
Question 35.5.4.Show that the pre-measure λ0we had, as well as the Borel measure ℬ(ℝn),are both σ-finite.
Actually, for us, σ-finite is basically always going to be true, so you can more or less just
take it for granted.
Theorem 35.5.5 (Carathéodory extension theorem for pre-measures) Let μ0 be a pre-measure
on an algebra 𝒜0 of Ω, and let 𝒜 denote the σ-algebra generated by 𝒜0. Let 𝒜 cm, μcm be
as in ?? . Then:
(a)
The restriction of μcm to 𝒜 gives a measure μ extending μ0.
(b)
If μ0 was σ-finite, then μ is the unique extension of μ0 to 𝒜 .
(c)
If μ0 was σ-finite, then μcm is the completion of μ, hence the unique extension of
μ0 to 𝒜 cm.
Here is the updated table, with comments if μ0 was indeed σ-finite.
Construct order
Notes
2Ω
μ∗
Step 2
μ∗ is outer measure obtained from μ0
𝒜 cm
μcm
Step 3
(𝒜 cm,μcm)
is completion (𝒜 ,μ),
𝒜 cm defined as μ∗-measurable sets
𝒜
μ
Step 4
𝒜 defined as σ-alg. generated by 𝒜0
𝒜0
μ0
Step 1
μ0 is a pre-measure
And here is the table for Ω = ℝn, with Borel and Lebesgue in it.
Construct order
Notes
2ℝn
λ∗
Step 2
λ∗ is Lebesgue outer measure
ℒ(ℝn)
λ
Step 3
Lebesgue σ-algebra, completion of Borel one
ℬ(ℝn)
μ
Step 4
Borel σ-algebra, generated by ℒ0
ℒ0
λ0
Step 1
Define pre-measure on rectangles
Going down one row of the table corresponds to restriction, while each of μ0→ μ → μcm is
a unique extension when μ0 is σ-finite.
Proof of?? . For (a): this is just ?? and ?? put together, combined with the
observation that 𝒜∗⊃ 𝒜0 and hence 𝒜∗⊃ 𝒜 . Parts (b) and (c) are more technical,
and omitted. □
35.6From now on, we assume the Borel measure
35.7A few harder problems to think about
Problem 35A† (Constructing outer measures from arbitrary ρ). For a set Ω, let ℰ be
any subset of 2Ω and let ρ: ℰ→ [0,+∞] be any function. Prove that
is an outer measure.
Problem 35B (The insane scientist).Let Ω = ℝ2, and let ℰ be the set of (non-degenerate)
rectangles. Let ρ(E) = 1 for every rectangle E ∈ℰ. Ignoring my advice, the insane scientist
uses ρ to construct an outer measure μ∗, as in ?? .
(a)
Find μ∗(S) for each subset S of ℝ2.
(b)
Which sets are μ∗-measurable?
You should find that no rectangle is μ∗-measurable, unsurprisingly foiling the scientist.
Problem 35C.A function f : ℝ → ℝ is continuous. Must f be
measurable with respect to the Lebesgue measure on ℝ?
36Lebesgue integration
On any measure space (Ω, 𝒜 ,μ) we can then, for a function f : Ω → [0,∞] define an
integral
This integral may be +∞ (even if f is finite). As the details of the construction won’t
matter for us later on, we will state the relevant definitions, skip all the proofs, and also state
all the properties that we actually care about. Consequently, this chapter will be quite
short.
36.1The definition
The construction is done in four steps.
Definition 36.1.1.If A is a measurable set of Ω, then the indicator function1A: Ω → ℝ is defined by
Step 1 (Indicator functions) — For an indicator function, we require
(which may be infinite).
We extend this linearly now for nonnegative functions which are sums of indicators: these
functions are called simple functions.
Step 2 (Simple functions) — Let A1, …, An be a finite collection of measurable sets. Let
c1, …, cn be either nonnegative real numbers or +∞. Then we define
If ci = ∞ and μ(Ai) = 0, we treat ciμ(Ai) = 0.
One can check the resulting sum does not depend on the representation of the simple
function as ∑ci1Ai. In particular, it is compatible with the previous step.
Conveniently, this is already enough to define the integral for f : Ω → [0,+∞]. Note that
[0,+∞] can be thought of as a topological space where we add new open sets (a,+∞] for each
real number a to our usual basis of open intervals. Thus we can equip it with the Borel
sigma-algebra.1
Step 3 (Nonnegative functions) — For each measurable function f : Ω → [0,+∞], let
where the supremum is taken over all simple s such that 0 ≤ s ≤ f. As before, this
integral may be +∞.
One can check this is compatible with the previous definitions. At this point, we introduce
an important term.
Definition 36.1.2.A measurable (nonnegative) function f : Ω → [0,+∞] is
absolutely integrable or just integrable if ∫Ωf dμ < ∞.
Warning: I find “integrable” to be really confusing terminology. Indeed, every measurable
function from Ω to [0,+∞] can be assigned a Lebesgue integral, it’s just that this integral may
be +∞. So the definition is far more stringent than the name suggests. Even constant
functions can fail to be integrable:
Example 36.1.3 (We really should call it “finitely integrable”) The constant function 1
is not integrable on ℝ, since ∫ℝ1 dμ = μ(ℝ) = +∞.
For this reason, I will usually prefer the term “integrable”. (If it were up to me, I would call
it “finitely integrable”, and usually do so privately.)
Finally, this lets us integrate general functions.
Definition 36.1.4.In general, a measurable function f : Ω → [−∞,∞] is absolutelyintegrable or just integrable if |f| is.
Since we’ll be using the first word, this is easy to remember: “absolutely integrable” requires
taking absolute values.
Step 4 (Absolutely integrable functions) — If f : Ω → [−∞,∞] is absolutely integrable, then
we define
f+(x)
= max
f−(x)
= min
and set
which in particular is finite.
You may already start to see that we really like nonnegative functions: with the theory of
measures, it is possible to integrate them, and it’s even okay to throw in +∞’s everywhere.
But once we start dealing with functions that can be either positive or negative, we have to
start adding finiteness restrictions — actually essentially what we’re doing is splitting
the function into its positive and negative part, requiring both are finite, and then
integrating.
To finish this section, we state for completeness some results that you probably could have
guessed were true. Fix Ω = (Ω, 𝒜 ,μ), and let f and g be measurable real-valued functions
such that f(x) = g(x) almost everywhere.
(Almost-everywhere preservation) The function f is absolutely integrable if and
only if g is, and if so, their Lebesgue integrals match.
(Additivity) If f and g are absolutely integrable then
The “absolutely integrable” hypothesis can be dropped if f and g are nonnegative.
(Scaling) If f is absolutely integrable and c ∈ ℝ then cf is absolutely integrable
and
The “absolutely integrable” hypothesis can be dropped if f is nonnegative and
c > 0.
(Monotoncity) If f and g are absolutely integrable and f ≤ g, then
The “absolutely integrable” hypothesis can be dropped if f and g are nonnegative.
There are more famous results like monotone/dominated convergence that are also true, but we
won’t state them here as we won’t really have a use for them in the context of probability.
(They appear later on in a bonus chapter.)
36.2Relation to Riemann integrals (or: actually computing Lebesgue integrals)
For closed intervals, this actually just works out of the box.
Theorem 36.2.1 (Lebesgue integral generalizes Riemann integral) Let f : [a,b] → ℝ be a
Riemann integrable function (where [a,b] is equipped with the Borel measure). Then f
is also Lebesgue integrable and the integrals agree:
Thus in practice, we do all theory with Lebesgue integrals (they’re nicer), but when we
actually need to compute ∫[1,4]x2dμ we just revert back to our usual antics with the
Fundamental Theorem of Calculus.
Example 36.2.2 (Integrating x2 over [1,4]) Reprising our old example:
This even works for improper integrals, if the functions are nonnegative. The statement is a
bit cumbersome to write down, but here it is.
Theorem 36.2.3 (Improper integrals are nice Lebesgue ones) Let f ≥ 0 be a nonnegative
continuous function defined on (a,b) ⊆ ℝ, possibly allowing a = −∞ or b = ∞. Then
where we allow both sides to be +∞ if f is not absolutely integrable.
The right-hand side makes sense since [a′,b′] ⊊ (a,b) is a compact interval on which f is
continuous. This means that improper Riemann integrals of nonnegative functions can just be
regarded as Lebesgue ones over the corresponding open intervals.
It’s probably better to just look at an example though.
Example 36.2.4 (Integrating 1∕ on (0,1)) For example, you might be familiar with
improper integrals like
(Note this appeared before as ?? .) In the Riemann integration situation, we needed the
limit as 𝜀 → 0+ since otherwise is not defined as a function [0,1] → ℝ. However, it
is a measurable nonnegative function (0,1) → [0,+∞], and hence
If f is not nonnegative, then all bets are off. Indeed ?? is the famous counterexample.
36.3A few harder problems to think about
Problem 36A⋆ (The indicator of the rationals). Take the indicator function
1ℚ : ℝ →{0,1}⊆ ℝ for the rational numbers.
(a)
Prove that 1ℚ is not Riemann integrable.
(b)
Show that ∫ℝ1ℚ exists and determine its value — the one you expect!
Problem 36B† (An improper Riemann integral with sign changes). Define
f : (1,∞) → ℝ by f(x) = . Show that f is not absolutely integrable, but that the
improper Riemann integral
nonetheless exists.
37Swapping order with Lebesgue integrals
37.1Motivating limit interchange
Prototypical example for this section:1ℚis good!
One of the issues with the Riemann integral is that it behaves badly with respect to
convergence of functions, and the Lebesgue integral deals with this. This is therefore often
given as a poster child or why the Lebesgue integral has better behaviors than the Riemann
one.
We technically have already seen this: consider the indicator function 1ℚ, which is not
Riemann integrable by ?? . But we can readily compute its Lebesgue integral over [0,1],
as
since it is countable.
This could be thought of as a failure of convergence for the Riemann integral.
Example 37.1.1 (1ℚ is a limit of finitely supported functions) We can define the
sequence of functions g1, g2, … by
Then each gn is piecewise continuous and hence Riemann integrable on [0,1] (with
integral zero), but limn→∞gn = 1ℚ is not.
The limit here is defined in the following sense:
Definition 37.1.2.Let f and f1,f2,…: Ω → ℝ be a sequence of functions. Suppose
that for each ω ∈ Ω, the sequence
converges to f(ω). Then we say (fn)nconverges pointwise to the limit f, written
limn→∞fn = f.
We can define liminf n→∞fn and limsupn→∞fn similarly.
This is actually a fairly weak notion of convergence, for example:
Exercise 37.1.3 (Witch’s hat).Find a sequence of continuous function on [−1,1] →ℝ whichconverges pointwise to the function f given by
This is why when thinking about the Riemann integral it is commonplace to work with
stronger conditions like “uniformly convergent” and the like. However, with the Lebesgue
integral, we can mostly not think about these!
37.2Overview
The three big-name results for exchanging pointwise limits with Lebesgue integrals is:
Fatou’s lemma: the most general statement possible, for any nonnegative
measurable functions.
Monotone convergence: “increasing limits” just work.
Dominated convergence (actually Fatou-Lebesgue): limits that are not too big
(bounded by some absolutely integrable function) just work.
37.3Fatou’s lemma
Without further ado:
Lemma 37.3.1 (Fatou’s lemma) Let f1,f2,…: Ω → [0,+∞] be a sequence of nonnegative
measurable functions. Then liminf n: Ω → [0,+∞] is measurable and
Here we allow either side to be +∞.
Notice that there are no extra hypothesis on fn other than nonnegative: which
makes this quite surprisingly versatile if you ever are trying to prove some general
result.
37.4Everything else
The big surprise is how quickly all the “big-name” theorem follows from Fatou’s lemma. Here
is the so-called “monotone convergence theorem”.
Corollary 37.4.1 (Monotone convergence theorem) Let f and f1,f2,…: Ω → [0,+∞] be a
sequence of nonnegative measurable functions such that limnfn = f and fn(ω) ≤ f(ω)
for each n. Then f is measurable and
Here we allow either side to be +∞.
Proof.We have
∫Ωf dμ
= ∫Ωdμ
≤ liminf n→∞∫Ωfndμ
≤ limsupn→∞∫Ωfndμ
≤∫Ωf dμ
where the first ≤ is by Fatou lemma, and the second by the fact that ∫Ωfn≤∫Ωf for every
n. This implies all the inequalities are equalities and we are done. □
Remark 37.4.2 (The monotone convergence theorem does not require monotonicity!) — In the literature it is much more common to see the hypothesis f1(ω) ≤ f2(ω) ≤≤f(ω) rather than just fn(ω) ≤ f(ω) for all n, which is where the theorem gets its
name. However as we have shown this hypothesis is superfluous! This is pointed out in
https://mathoverflow.net/a/296540/70654, as a response to a question entitled “Do
you know of any very important theorems that remain unknown?”.
Example 37.4.3 (Monotone convergence gives 1ℚ) This already implies ?? . Letting gn
be the indicator function for ℤ as described in that example, we have gn≤ 1ℚ and
limn→∞gn(x) = 1ℚ(x), for each individual x. So since ∫[0,1]gndμ = 0 for each n, this
gives ∫[0,1]1ℚ = 0 as we already knew.
The most famous result, though is the following.
Corollary 37.4.4 (Fatou–Lebesgue theorem) Let f and f1,f2,…: Ω → ℝ be a sequence of
measurable functions. Assume that g: Ω → ℝ is an absolutely integrable function for which
|fn(ω)|≤|g(ω)| for all ω ∈ Ω. Then the inequality
∫Ωdμ
≤ liminf n→∞
≤ limsupn→∞≤∫Ωdμ.
Proof.There are three inequalities:
The first inequality follows by Fatou on g + fn which is nonnegative.
The second inequality is just liminf ≤ limsup. (This makes the theorem statement
easy to remember!)
The third inequality follows by Fatou on g − fn which is nonnegative. □
Exercise 37.4.5.Where is the fact that g is absolutely integrable used in this proof?
Corollary 37.4.6 (Dominated convergence theorem) Let f1,f2,…: Ω → ℝ be a sequence
of measurable functions such that f = limn→∞fn exists. Assume that g: Ω → ℝ is an
absolutely integrable function for which |fn(ω)|≤|g(ω)| for all ω ∈ Ω. Then
Proof.If f(ω) = limn→∞fn(ω), then f(ω) = liminf n→∞fn(ω) = limsupn→∞fn(ω). So
all the inequalities in the Fatou-Lebesgue theorem become equalities, since the leftmost
and rightmost sides are equal. □
Note this gives yet another way to verify ?? . In general, the dominated convergence
theorem is a favorite cliché for undergraduate exams, because it is easy to create questions
for it. Here is one example showing how they all look.
Example 37.4.7 (The usual Lebesgue dominated convergence examples) Suppose one
wishes to compute
then one starts by observing that the inner term is bounded by the absolutely integrable
function x−1∕2. Therefore it equals
∫(0,1) limn→∞dx
= ∫(0,1)dx
= ∫(0,1)dx = .
37.5Fubini and Tonelli
37.6A few harder problems to think about
38Bonus: A hint of Pontryagin duality
In this short chapter we will give statements about how to generalize our Fourier analysis (a
bonus chapter ?? ) to a much wider class of groups G.
38.1LCA groups
Prototypical example for this section:𝕋, ℝ.
Earlier we played with ℝ, which is nice because in addition to being a topological space, it is
also an abelian group under addition. These sorts of objects which are both groups and spaces
have a name.
Definition 38.1.1.A group G is a topological group is a
Hausdorff1
topological space equipped also with a group operation (G,⋅), such that both maps
G × G
→ G by (x,y)xy
G
→ G by xx−1
are continuous.
For our Fourier analysis, we need some additional conditions.
Definition 38.1.2.A locally compact abelian (LCA) groupG is one for which
the group operation is abelian, and moreover the topology is locally compact: for every
point p of G, there exists a compact subset K of G such that K ∋ p, and K contains
some open neighborhood of p.
Our previous examples all fall into this category:
Example 38.1.3 (Examples of locally compact abelian groups)
Any finite group Z with the discrete topology is LCA.
The circle group 𝕋 is LCA and also in fact compact.
The real numbers ℝ are an example of an LCA group which is not compact.
These conditions turn out to be enough for us to define a measure on the space G. The
relevant theorem, which we will just quote:
Theorem 38.1.4 (Haar measure) Let G be a locally compact abelian group. We regard
it as a measurable space using its Borel σ-algebra ℬ(G). There exists a measure
μ: ℬ(G) → [0,∞], called the Haar measure, satisfying the following properties:
μ(gS) = μ(S) for every g ∈ G and measurable S. That means that μ is
“translation-invariant” under translation by G.
μ(K) is finite for any compact set K.
if S is measurable, then μ(S) = inf .
if U is open, then μ(U) = sup.
Moreover, it is unique up to scaling by a positive constant.
Remark 38.1.5 — Note that if G is compact, then μ(G) is finite (and positive). For
this reason the Haar measure on a LCA group G is usually normalized so μ(G) = 1.
For this chapter, we will only use the first two properties at all, and the other two are just
mentioned for completeness. Note that this actually generalizes the chapter where we
constructed a measure on ℬ(ℝn), since ℝn is an LCA group!
So, in short: if we have an LCA group, we have a measure μ on it.
38.2The Pontryagin dual
Now the key definition is:
Definition 38.2.1.Let G be an LCA group. Then its Pontryagin dual is the abelian
group
The maps ξ are called characters. It can be itself made into an LCA group.2
Example 38.2.2 (Examples of Pontryagin duals)
ℤ𝕋, since group homomorphisms ℤ → 𝕋 are determined by the image of 1.
𝕋ℤ. The characters are given by 𝜃n𝜃 for n ∈ ℤ.
ℝℝ. This is because a nonzero continuous homomorphism ℝ → S1 is determined
by the fiber above 1 ∈ S1. (Algebraic topologists might see covering projections
here.)
ℤ∕nℤℤ∕nℤ, characters ξ being determined by the image ξ(1) ∈ 𝕋.
G × HG×H.
Exercise 38.2.3 (ZZ, for those who read?? ).If Z is a finite abelian group, show thatZZ, using the results of the previous example. You may now recognize that the bilinear form⋅:Z × Z →𝕋 is exactly a choice of isomorphism Z →Z. It is not “canonical”.
True to its name as the dual, and in analogy with (V∨)∨V for vector spaces V , we
have:
Theorem 38.2.4 (Pontryagin duality theorem) For any LCA group G, there is an
isomorphism
The compact case is especially nice.
Proposition 38.2.5 (G compact⇔Gdiscrete) Let G be an LCA group. Then G is
compact if and only if G is discrete.
Proof.?? . □
38.3The orthonormal basis in the compact case
Let G be a compact LCA group, and work with its Haar measure. We may now let L2(G) be
the space of square-integrable functions to ℂ, i.e.
Thus we can equip it with the inner form
In that case, we get all the results we wanted before:
Theorem 38.3.1 (Characters ofGforms an orthonormal basis) Assume G is LCA and
compact (so G is discrete). Then the characters
form an orthonormal basis of L2(G). Thus for each f ∈ L2(G) we have
where
The sum ∑ξ∈G makes sense since G is discrete. In particular,
Letting G = Z for a finite group G gives “Fourier transform on finite groups”.
The special case G = ℤ∕nℤ has its own Wikipedia page: the “discrete-time Fourier
transform”.
Letting G = 𝕋 gives the “Fourier series” earlier.
38.4The Fourier transform of the non-compact case
If G is LCA but not compact, then Theorem 38.3.1 becomes false. On the other hand, it’s still
possible to define G. We can then try to write the Fourier coefficients anyways:
let
for ξ ∈G and f : G → ℂ. The results are less fun in this case, but we still have, for
example:
Theorem 38.4.1 (Fourier inverison formula in the non-compact case) Let μ be a Haar
measure on G. Then there exists a unique Haar measure ν on G (called the dualmeasure) such that: whenever f ∈ L1(G) and f∈ L1(G), we have
for almost all x ∈ G (with respect to μ). If f is continuous, this holds for all x.
So while we don’t have the niceness of a full inner product from before, we can still in some
situations at least write f as integral in sort of the same way as before.
In particular, they have special names for a few special G:
We summarize our various flavors of Fourier analysis from the previous sections in the
following table. In the first part G is compact, in the second half G is not.
You might notice that the various names are awful. This is part of the reason I got
confused as a high school student: every type of Fourier series above has its own Wikipedia
article. If it were up to me, we would just use the term “G-Fourier transform”, and that would
make everyone’s lives a lot easier.
38.6A few harder problems to think about
Problem 38A.If G is compact, so G is discrete, describe the dual measure ν.
Problem 38B. Show that an LCA group G is compact if and only if G is discrete. (You will
need the compact-open topology for this.)
Part XI Probability (TO DO)
39Random variables (TO DO)
Having properly developed the Lebesgue measure and the integral on it, we can now proceed
to develop random variables.
39.1Random variables
With all this set-up, random variables are going to be really quick to define.
Definition 39.1.1.A (real) random variableX on a probability space Ω = (Ω, 𝒜 ,μ)
is a measurable function X : Ω → ℝ, where ℝ is equipped with the Borel σ-algebra.
In particular, addition of random variables, etc. all makes sense, as we can just add. Also, we
can integrate X over Ω, by previous chapter.
Definition 39.1.2 (First properties of random variables).Given a random variable X,
the expected value of X is defined by the Lebesgue integral
Confusingly, the letter μ is often used for expected values.
The kth moment of X is defined as 𝔼[Xk], for each positive integer k ≥ 1. The
variance of X is then defined as
Question 39.1.3.Show that 1Ais a random variable (just check that it is Borel measurable),and its expected value is μ(A).
An important property of expected value you probably already know:
Theorem 39.1.4 (Linearity of expectation) If X and Y are random variables on Ω then
Note that X and Y do not have to be “independent” here: a notion we will define
shortly.
39.2Distribution functions
39.3Examples of random variables
39.4Characteristic functions
39.5Independent random variables
39.6A few harder problems to think about
Problem 39A (Equidistribution).Let X1, X2, …be i.i.d. uniform random variables on
[0,1]. Show that almost surely the Xi are equidistributed, meaning that
holds for almost all choices of ω.
Problem 39B (Side length of triangle independent from median).Let X1, Y1, X2,
Y2, X3, Y3 be six independent standard Gaussians. Define triangle ABC in the Cartesian
plane by A = (X1,Y1), B = (X2,Y2), C = (X3,Y3). Prove that the length of side BC
is independent from the length of the A-median.
40Large number laws (TO DO)
40.1Notions of convergence
40.1.iAlmost sure convergence
Definition 40.1.1.Let X, Xn be random variables on a probability space Ω. We say
Xnconverges almost surely to X if
This is a very strong notion of convergence: it says in almost every world, the values of Xn
converge to X. In fact, it is almost better for me to give a non-example.
Example 40.1.2 (Non-example of almost sure convergence) Imagine an immortal skeleton archer is practicing shots, and on the nth shot, he scores
a bulls-eye with probability 1 − (which tends to 1 because the archer improves over
time). Let Xn∈{0,1,…,10} be the score of the nth shot.
Although the skeleton is gradually approaching perfection, there are almost no worlds
in which the archer misses only finitely many shots: that is
40.1.iiConvergence in probability
Therefore, for many purposes we need a weaker notion of convergence.
Definition 40.1.3.Let X, Xn be random variables on a probability space Ω. We say
Xnconverges in probability to X if if for every 𝜀 > 0 and δ > 0, we have
for n large enough (in terms of 𝜀 and δ).
In this sense, our skeleton archer does succeed: for any δ > 0, if n > δ−1 then the skeleton
archer does hit a bulls-eye in a 1 −δ fraction of the worlds. In general, you can think of this as
saying that for any δ > 0, the chance of an 𝜀-anomaly event at the nth stage eventually drops
below δ.
Remark 40.1.4 — To mask δ from the definition, this is sometimes written instead as:
for all 𝜀
I suppose it doesn’t make much difference, though I personally don’t like the asymmetry.
40.1.iiiConvergence in law
40.2A few harder problems to think about
Problem 40A (Quantifier hell).In the definition of convergence in
probability suppose we allowed δ = 0 (rather than δ > 0). Show that the modified definition is
equivalent to almost sure convergence.
Problem 40B (Almost sure convorgence is not topologizable).Consider the space of all
random variables on Ω = [0,1]. Prove that it’s impossible to impose a metric on this space
which makes the following statement true:
A sequence X1, X2, …, of converges almost surely to X if and only if Xi
converge to X in the metric.
41Stopped martingales (TO DO)
41.1How to make money almost surely
We now take our newfound knowledge of measure theory to a casino.
Here’s the most classical example that shows up: a casino lets us play a game where we can
bet any amount of on a fair coin flip, but with bad odds: we win $n if the coin is heads, but
lose $2n if the coin is tails, for a value of n of our choice. This seems like a game that no one in
their right mind would want to play.
Well, if we have unbounded time and money, we actually can almost surely make a
profit.
Example 41.1.1 (Being even greedier than 18th century France) In the game above, we start
by betting $1.
If we win, we leave having made $1.
If we lose, we then bet $10 instead, and
If we win, then we leave having made $10 − $2 = $8, and
If we lose then we bet $100 instead, and
If we win, we leave having made $1000 − $20 − $2 = $978, and
If we lose then we bet $1000 instead, and so on…
Since the coin will almost surely show heads eventually, we make money whenever that
happens. In fact, the expected amount of time until a coin shows heads is only 2 flips! What
could go wrong?
This chapter will show that under sane conditions such as “finite time” or “finite money”,
one cannot actually make money in this way — the optional stopping theorem. This will give
us an excuse to define conditional probabilities, and then talk about martingales (which
generalize the fair casino).
Once we realize that trying to extract money from Las Vegas is a lost cause, we will stop
gambling and then return to solving math problems, by showing some tricky surprises, where
problems that look like they have nothing to do with gambling can be solved by considering a
suitable martingale.
In everything that follows, Ω = (Ω, 𝒜 ,μ) is a probability space.
41.2Sub-σ-algebras and filtrations
Prototypical example for this section:σ-algebra generated by a random variable, and coin flipfiltration.
We considered our Ω as a space of worlds, equipped with a σ-algebra 𝒜 that lets us
integrate over Ω. However, it is a sad fact of life that at any given time, you only know partial
information about the world. For example, at the time of writing, we know that the world did
not end in 2012 (see https://en.wikipedia.org/wiki/2012_phenomenon), but the fate of
humanity in future years remains at slightly uncertain.
Let’s write this measure-theoretically: we could consider
Ω
= A ⊔ B
A
=
B
= .
We will assume that A and B are measurable sets, that is, A,B ∈ 𝒜 . That means we
could have good fun arguing about what the values of μ(A) and μ(B) should be (“a
priori probability that the world ends in 2012”), but let’s move on to a different silly
example.
We will now introduce a new notion that we will need when we define conditional
probabilities later.
Definition 41.2.1.Let Ω = (Ω, 𝒜 ,μ) be a probability space. A sub-σ-algebraℱ on
Ω is exactly what it sounds like: a σ-algebra ℱ on the set Ω such that each A ∈ ℱ is
measurable (i.e., ℱ ⊆ 𝒜 ).
The motivation is that ℱ is the σ-algebra of sets which let us ask questions about some
piece of information. For example, in the 2012 example we gave above, we might take
ℱ = {∅,A,B,Ω}, which are the sets we care about if we are thinking only about
2012.
Here are some more serious examples.
Example 41.2.2 (Examples of sub-σ-algebras)
(a)
Let X : Ω →{0,1,2} be a random variable taking on one of three values. If we’re
interested in X then we could define
A
= {ω: X(ω) = 1}
B
= {ω: X(ω) = 2}
C
= {ω: X(ω) = 3}
then we could write
This is a sub-σ-algebra on ℱ that lets us ask questions about X like “what is the
probability X≠3”, say.
(b)
Now suppose Y : Ω → [0,1] is another random variable. If we are interested in Y , the ℱ
that captures our curiosity is
You might notice a trend here which we formalize now:
Definition 41.2.3.Let X : Ω → ℝ be a random variable. The sub-σ-algebragenerated by X is defined by
If X1, …is a sequence (finite or infinite) of random variables, the sub-σ-algebra
generated by them is the smallest σ-algebra which contains σ(Xi) for each i.
Finally, we can put a lot of these together — since we’re talking about time, we learn more
as we grow older, and this can be formalized.
Definition 41.2.4.A filtration on Ω = (Ω, 𝒜 ,μ) is a nested sequence1
of sub-σ-algebras on Ω.
Example 41.2.5 (Filtration) Suppose you’re bored in an infinitely long class and start
flipping a fair coin to pass the time. (Accordingly, we could let Ω = {H,T}∞ consist of infinite
sequences of heads H and tails T.) We could let ℱn denote the sub-σ-algebra generated by
the values of the first n coin flips. So:
ℱ0 = {∅,Ω},
ℱ1 = {∅,first flip H,first flip T,Ω},
ℱ2 = {∅,first flips HH,second flip T,Ω,first flip and second flip differ,…}.
and so on, with ℱn being the measurable sets “determined” only by the first n
coin flips.
Exercise 41.2.6.In the previous example, compute the cardinality |ℱn| for each integer n.
41.3Conditional expectation
Prototypical example for this section:𝔼(X∣X + Y ) for X and Y distributed over
[0,1].
We’ll need the definition of conditional probability to define a martingale, but this turns
out to be surprisingly tricky. Let’s consider the following simple example to see
why.
Example 41.3.1 (Why high-school methods aren’t enough here) Suppose we have two
independent random variables X, Y distributed uniformly over [0,1] (so we may as well take
Ω = [0,1]2). We might try to ask the question:
“what is the expected value of X given that X + Y = 0.6”?
Intuitively, we know the answer has to be 0.3. However, if we try to write down a definition,
we quickly run into trouble. Ideally we want to say something like
The problem is that S is a set of measure zero, so we quickly run into , meaning a definition
of this shape will not work out.
The way that this is typically handled in measure theory is to use the notion of
sub-σ-algebra that we defined. Let ℱ be a sub-σ-algebra which captures the information This
means we create a function assigning the “conditional expectation” to every point ω ∈ Ω,
which is measurable with respect to ℱ .
Proposition 41.3.2 (Conditional expectation definition) Let X : Ω → ℝ be an absolutelyintegrable random variable (meaning 𝔼[|X|] < ∞) over a probability space Ω, and let ℱ
be a sub-σ-algebra on it.
Then there exists a function η: Ω → ℝ satisfying the following two properties:
η is ℱ -measurable (that is, measurable as a function (Ω, ℱ ,μ) → ℝ); and
for any set A ∈ ℱ we have 𝔼[η ⋅1A] = 𝔼[X ⋅1A].
Moreover, this random variable is unique up to almost sureness.
Proof.Omitted, but relevant buzzword used is “Radon-Nikodym derivative”. □
Definition 41.3.3.Let η be as in the previous proposition.
We denote η by 𝔼(X∣ℱ ) and call it the conditional expectation of X with
respect to ℱ .
If Y is a random variable then 𝔼(X∣Y ) denotes 𝔼(X∣σ(Y )), i.e. the conditional
expectation of X with respect to the σ-algebra generated by Y .
More fine print:
Remark 41.3.4 (This notation is terrible) — The notation 𝔼(X∣ℱ ) is admittedly
confusing, since it is actually an entire function Ω → ℝ, rather than just a real number
like 𝔼[X]. For this reason I try to be careful to remember to use parentheses rather than
square brackets for conditional expectations; not everyone does this.
Abuse of Notation 41.3.5.In addition, when we write Y = 𝔼(X∣ℱ ), there is some
abuse of notation happening here since 𝔼(X∣ℱ ) is defined only up to some reasonable
uniqueness (i.e. up to measure zero changes). So this really means that “Y satisfies the
hypothesis of ?? ”, but this is so pedantic that no one bothers.
41.4Supermartingales
Prototypical example for this section:Visiting a casino is a supermartingale, assuming houseodds.
Definition 41.4.1.Let X0, X1, …be a sequence of random variables on a probability
space Ω, and let ℱ0⊆ ℱ1⊆ be a filtration.
Then (Xn)n≥0 is a supermartingale with respect to (ℱn)n≥0 if the following conditions
hold:
Xn is absolutely integrable for every n;
Xn is measurable with respect to ℱn; and
for each n = 1,2,… the inequality
holds for all ω ∈ Ω.
In a submartingale the inequality ≤ is replaced with ≥, and in a martingale it is
replaced by =.
Abuse of Notation 41.4.2 (No one uses that filtration thing anyways).We
will always take ℱn to be the σ-algebra generated by the previous variables X0, X1, …,
Xn−1, and do so without further comment. Nonetheless, all the results that follow hold
in the more general setting of a supermartingale with respect to some filtration.
We will prove all our theorems for supermartingales; the analogous versions for
submartingales can be obtained by replacing ≤ with ≥ everywhere (since Xn is a martingale
iff −Xn is a supermartingale) and for martingales by replacing ≤ with = everywhere (since
Xn is a martingale iff it is both a supermartingale and a submartingale).
Let’s give examples.
Example 41.4.3 (Supermartingales)
(a)
Random walks: an ant starts at the position 0 on the number line. Every minute,
it flips a fair coin and either walks one step left or one step right. If Xt is the position
at the tth time, then Xt is a martingale, because
(b)
Casino game: Consider a gambler using the strategy described at the beginning of
the chapter. This is a martingale, since every bet the gambler makes has expected
value 0.
(c)
Multiplying independent variables: Let X1, X2, …, be independent (not
necessarily identically distributed) integrable random variables with mean 1. Then
the sequence Y1, Y2 …defined by
is a martingale; as 𝔼(Yn∣Y1,…,Yn−1) = 𝔼[Yn] ⋅ Yn−1 = Yn−1.
(d)
Iterated blackjack: Suppose one shows up to a casino and plays infinitely many
games of blackjack. If Xt is their wealth at time t, then Xt is a supermartingale.
This is because each game has negative expected value (house edge).
Example 41.4.4 (Frivolous/inflamamtory example — real life is a supermartingale)
Let Xt be your happiness on day t of your life. Life has its ups and downs, so it is not
the case that Xt≤ Xt−1 for every t. For example, you might win the lottery one day.
However, on any given day, many things can go wrong (e.g. zombie apocalypse), and by
Murphy’s Law this is more likely than things going well. Also, as you get older, you have
an increasing number of responsibilities and your health gradually begins to deteriorate.
Thus it seems that
is a reasonable description of the future — in expectation, each successive day is slightly
worse than the previous one. (In particular, if we set Xt = −∞ on death, then as long
as you have a positive probability of dying, the displayed inequality is obviously true.)
Before going on, we will state without proof one useful result: if a martingale is bounded,
then it will almost certainly converge.
Theorem 41.4.5 (Doob’s martingale convergence theorem) Let X0, …be a
supermartingale on a probability space Ω such that
Then, there exists a random variable X∞: Ω → ℝ such that
41.5Optional stopping
Prototypical example for this section:Las Vegas.
In the first section we described how to make money almost surely. The key advantage the
gambler had was the ability to quit whenever he wanted (equivalently, an ability to control the
size of the bets; betting $0 forever is the same as quitting.) Let’s formalize a notion of
stopping time.
The idea is we want to define a function τ : Ω →{0,1,…,∞} such that
τ(ω) specifies the index after which we stop the martingale. Note that the decisions
to stop after time n must be made with only the information available at that time
— i.e., with respect to ℱn of the filtration.
Xτ∧n is the random value representing the value at time n of the stopped
martingale, where if n is after the stopping time, we just take it to be the our
currently value after we leave.
So for example in a world ω where we stopped at time 3, then Xτ∧0(ω) = X0(ω),
Xτ∧1(ω) = X1(ω), Xτ∧2(ω) = X2(ω), Xτ∧3(ω) = X3(ω), but then
since we have stopped — the value stops changing.
Xτ denotes the eventual value after we stop (or the limit X∞ if we never stop).
Here’s the compiled machine code.
Definition 41.5.1.Let ℱ0⊆ ℱ1⊆ be a filtration on a probability space Ω.
A stopping time is a function
with the property that for each integer n, the set
is ℱn-measurable (i.e., is in ℱn).
For each n ≥ 0 we define Xτ∧n: Ω → ℝ by
Finally, we let the eventual outcome be denoted by
We require that the “undefined” case occurs only for a set of measure zero (for
example, if ?? applies). Otherwise we don’t allow Xτ to be defined.
Proposition 41.5.2 (Stopped supermartingales are still supermartingales) Let X0, X1, …be
a supermartingale. Then the sequence
is itself a supermartingale.
Proof.We have almost everywhere the inequalities
𝔼
= 𝔼
= 𝔼 + 𝔼
= Xn−1 + 1τ(ω)=n−1⋅ 𝔼≤ Xn−1
as functions from Ω → ℝ. □
Theorem 41.5.3 (Doob’s optional stopping theorem) Let X0, X1, …be a supermartingale on a
probability space Ω, with respect to a filtration ℱ0⊆ ℱ1⊆. Let τ be a stopping time
with respect to this filtration. Suppose that any of the following hypotheses are true, for some
constant C:
(a)
Finite time: τ(ω) ≤ C for almost all ω.
(b)
Finite money: for each n ≥ 1, ≤ C for almost all ω.
(c)
Finite bets: we have 𝔼[τ] < ∞, and for each n ≥ 1, the conditional expectation
takes on values at most C for almost all ω ∈ Ω satisfying τ(ω) ≥ n.
Then Xτ is well-defined almost everywhere, and more importantly,
The last equation can be cheekily expressed as “the only winning move is not to
play”.
Proof. □
Exercise 41.5.4.Conclude that going to Las Vegas with the strategy described in the firstsection is a really bad idea. What goes wrong?
41.6Fun applications of optional stopping (TO DO)
We now give three problems which showcase some of the power of the results we have
developed so far.
41.6.iThe ballot problem
Suppose Alice and Bob are racing in an election; Alice received a votes total while Bob
received b votes total, and a > b. If the votes are chosen in random order, one could ask: what
is the probability that Alice never falls behind Bob in the election?
Proposition 41.6.1 (Ballot problem) This occurs with probability .
41.6.iiABRACADABRA
41.6.iiiUSA TST 2018
41.7A few harder problems to think about
Problem 41A (Examples of martingales).We give some more examples of martingales.
(a)
(Simple random walk) Let X1, X2, …be i.i.d. random variables which equal +1
with probability 1∕2, and −1 with probability 1∕2. Prove that
is a martingale.
(b)
(de Moivre’s martingale) Fix real numbers p and q such that p,q > 0 and
p + q = 1. Let X1, X2, …be i.i.d. random variables which equal +1 with probability
p, and −1 with probability q. Show that
is a martingale.
(c)
(Pólya’s urn) An urn contains one red and one blue marble initially. Every
minute, a marble is randomly removed from the urn, and two more marbles of the
same color are added to the urn. Thus after n minutes, the urn will have n + 2
marbles.
Let rn denote the fraction of marbles which are red. Show that rn is a martingale.
Problem 41B.A deck has 52 cards; of them 26 are red and 26 are black. The cards are
drawn and revealed one at a time. At any point, if there is at least one card remaining in the
deck, you may stop the dealer; you win if (and only if) the next card in the deck is red. If all
cards are dealt, then you lose. Across all possible strategies, determine the maximal
probability of winning.
Problem 41C (Wald’s identity).Let μ be a real number. Let X1, X2, …be independent
random variables on a probability space Ω with mean μ. Finally let τ : Ω →{1,2,…} be
a stopping time such that 𝔼[τ] < ∞, such that the event τ = n depends only on X1, …,
Xn.
Prove that
Problem 41D (Unbiased drunkard’s walk).An ant starts at 0 on a number line, and walks
left or right one unit with probability 1∕2. It stops once it reaches either −17 or
+8.
(a)
Find the probability it reaches +8 before −17.
(b)
Find the expected value of the amount of time it takes to reach either endpoint.
Problem 41E (Biased drunkard’s walk).Let 0 < p < 1 be a real number. An ant starts
at 0 on a number line, and walks left or right one unit with probability p. It stops once
it reaches either −17 or +8. Find the probability it reaches +8 first.
Problem 41F.The number 1 is written on a blackboard. Every minute, if the number a is
written on the board, it’s erased and replaced by a real number in the interval [0,2.01a]
selected uniformly at random. What is the probability that the resulting sequence of numbers
approaches 0?
Part XII Differential Geometry
42Multivariable calculus done correctly
As I have ranted about before, linear algebra is done wrong by the extensive use of matrices to
obscure the structure of a linear map. Similar problems occur with multivariable calculus, so
here I would like to set the record straight.
Since we are doing this chapter using morally correct linear algebra, it’s imperative you’re
comfortable with linear maps, and in particular the dual space V∨ which we will repeatedly
use.
In this chapter, all vector spaces have norms and are finite-dimensional over ℝ. So in
particular every vector space is also a metric space (with metric given by the norm), and we
can talk about open sets as usual.
42.1The total derivative
Prototypical example for this section:If f(x,y) = x2 + y2, then (Df)(x,y) = 2xe1∨ + 2ye2∨.
First, let f : [a,b] → ℝ. You might recall from high school calculus that for every point
p ∈ ℝ, we defined f′(p) as the derivative at the point p (if it existed), which we interpreted as
the slope of the “tangent line”.
That’s fine, but I claim that the “better” way to interpret the derivative at that point is as
a linear map, that is, as a function. If f′(p) = 1.5, then the derivative tells me that if
I move 𝜀 away from p then I should expect f to change by about 1.5𝜀. In other
words,
The derivative of f at p approximates f near p by a linear function.
What about more generally? Suppose I have a function like f : ℝ2→ ℝ, say
for concreteness or something. For a point p ∈ ℝ2, the “derivative” of f at p ought to
represent a linear map that approximates f at that point p. That means I want a linear map
T : ℝ2→ ℝ such that
for small displacements v ∈ ℝ2.
Even more generally, if f : U → W with U ⊆ V open (in the V metric as usual), then
the derivative at p ∈ U ought to be so that
(We need U open so that for small enough v, p + v ∈ U as well.) In fact this is exactly what
we were doing earlier with f′(p) in high school.
Image derived from [?]
The only difference is that, by an unfortunate coincidence, a linear map ℝ → ℝ can be
represented by just its slope. And in the unending quest to make everything a number so that
it can be AP tested, we immediately forgot all about what we were trying to do
in the first place and just defined the derivative of f to be a number instead of a
function.
The fundamental idea of Calculus is the local approximation of functions by linearfunctions. The derivative does exactly this.
Jean Dieudonné as quoted in [?] continues:
In the classical teaching of Calculus, this idea is immediately obscured
by the accidental fact that, on a one-dimensional vector space, there is a
one-to-one correspondence between linear forms and numbers, and therefore
the derivative at a point is defined as a number instead of a linear form. This
slavish subservience to the shibboleth of numerical interpretationat any cost becomes much worse . . .
So let’s do this right. The only thing that we have to do is say what “≈” means, and for this
we use the norm of the vector space.
Definition 42.1.1.Let U ⊆ V be open. Let f : U → W be a continuous function, and
p ∈ U. Suppose there exists a linear map T : V → W such that
Then T is the total derivative of f at p. We denote this by (Df)p, and say f is
differentiable at p.
If (Df)p exists at every point, we say f is differentiable.
Question 42.1.2.Check if that V = W = ℝ, this is equivalent to the single-variable definition.(What are the linear maps from V to W?)
Example 42.1.3 (Total derivative of f(x,y) = x2 + y2) Let V = ℝ2 with standard
basis e1, e2 and let W = ℝ, and let f = x2 + y2. Let p = ae1 + be2. Then,
we claim that
Here, the notation e1∨ and e2∨ makes sense, because by definition (Df)p∈ V∨: these are
functions from V to ℝ!
Let’s check this manually with the limit definition. Set v = xe1 + ye2, and note that the
norm on V is V = while the norm on W is just the absolute value
W = . Then we compute
=
= =
→ 0
as → 0. Thus, for p = ae1 + be2 we indeed have (Df)p = 2a ⋅ e1∨ + 2b ⋅ e2∨.
Remark 42.1.4 — As usual, differentiability implies continuity.
Remark 42.1.5 — Although U ⊆ V , it might be helpful to think of vectors from U
and V as different types of objects (in particular, note that it’s possible for 0VU).
The vectors in U are “inputs” on our space while the vectors coming from V are “small
displacements”. For this reason, I deliberately try to use p ∈ U and v ∈ V when possible.
42.2The projection principle
Before proceeding I need to say something really important.
Theorem 42.2.1 (Projection principle) Let U be an open subset of the vector space
V . Let W be an n-dimensional real vector space with basis w1,…,wn. Then there
is a bijection between continuous functions f : U → W and n-tuples of continuous
f1,f2,…,fn : U → ℝ by projection onto the ith basis element, i.e.
Proof.Obvious. □
The theorem remains true if one replaces “continuous” by “differentiable”, “smooth”,
“arbitrary”, or most other reasonable words. Translation:
To think about a function f : U → ℝn, it suffices to think about each coordinateseparately.
For this reason, we’ll most often be interested in functions f : U → ℝ. That’s why the dual
space V∨ is so important.
42.3Total and partial derivatives
Prototypical example for this section:If f(x,y) = x2 + y2, then (Df) : (x,y)2x⋅e1∨ + 2y ⋅e2∨,and = 2x, = 2y.
Let U ⊆ V be open and let V have a basis e1, …, en. Suppose f : U → ℝ is a function which
is differentiable everywhere, meaning (Df)p∈ V∨ exists for every p. In that case, one can
consider Df as itself a function:
Df : U
→ V∨
p
(Df)p.
This is a little crazy: to every point in U we associate a function in V∨. We say Df is the
total derivative of f, to reflect how much information we’re dealing with. We say (Df)p is
the total derivative at p.
Let’s apply the projection principle now to Df. Since we picked a basis e1, …, en of V , there
is a corresponding dual basis e1∨, e2∨, …, en∨. The Projection Principle tells us that Df can
thus be thought of as just n functions, so we can write
In fact, we can even describe what the ψi are.
Definition 42.3.1.The ithpartial derivative of f : U → ℝ, denoted
is defined by
You can think of it as “f′ along ei”.
Question 42.3.2.Check that if Df exists, then
Remark 42.3.3 — Of course you can write down a definition of for any v (rather
than just the ei).
From the above remarks, we can derive that
and so given a basis of V , we can think of Df as just the n partials.
Remark 42.3.4 — Keep in mind that each is a function from U to the reals. That
is to say,
Example 42.3.5 (Partial derivatives of f(x,y) = x2 + y2) Let f : ℝ2→ ℝ by
(x,y)x2 + y2. Then in our new language,
Thus the partials are
With all that said, I haven’t really said much about how to find the total derivative itself.
For example, if I told you
you might want to be able to compute Df without going through that horrible limit
definition I told you about earlier.
Fortunately, it turns out you already know how to compute partial derivatives, because you
had to take AP Calculus at some point in your life. It turns out for most reasonable functions,
this is all you’ll ever need.
Theorem 42.3.6 (Continuous partials implies differentiable) Let U ⊆ V be open and pick
any basis e1,…,en. Let f : U → ℝ and suppose that is defined for each i and moreover
is continuous. Then f is differentiable and Df is given by
Proof.Not going to write out the details, but…given v = t1e1 + + tnen, the idea is to
just walk from p to p + t1e1, p + t1e1 + t2e2, …, up to p + t1e1 + t2e2 + + tnen = p + v,
picking up the partial derivatives on the way. Do some calculation. □
Remark 42.3.7 — The continuous condition cannot be dropped. The function
is the classic counterexample – the total derivative Df does not exist at zero, even
though both partials do.
Example 42.3.8 (Actually computing a total derivative) Let f(x,y) = xsiny + x2y4.
Then
(x,y)
= siny + y4⋅ 2x
(x,y)
= xcosy + x2⋅ 4y3.
So ?? applies, and Df = e1∨ + e2∨, which I won’t bother to write out.
The example f(x,y) = x2 + y2 is the same thing. That being said, who cares about
xsiny + x2y4 anyways?
42.4(Optional) A word on higher derivatives
Let U ⊆ V be open, and take f : U → W, so that Df : U → Hom(V,W).
Well, Hom(V,W) can also be thought of as a normed vector space in its own right: it turns
out that one can define an operator norm on it by setting
So Hom(V,W) can be thought of as a normed vector space as well. Thus it makes sense to
write
which we abbreviate as D2f. Dropping all doubt and plunging on,
I’m sorry. As consolation, we at least know that Hom(V,W)V∨⊗W in a natural way, so
we can at least condense this to
rather than writing a bunch of Hom’s.
Remark 42.4.1 — If k = 2, W = ℝ, then D2f(v) ∈ (V∨)⊗2, so it can be represented
as an n × n matrix, which for some reason is called a Hessian.
The most important property of the second derivative is that
Theorem 42.4.2 (Symmetry of D2f) Let f : U → W with U ⊆ V open. If (D2f)p exists
at some p ∈ U, then it is symmetric, meaning
I’ll just quote this without proof (see e.g. [?, §5, theorem 16]), because double derivatives
make my head spin. An important corollary of this theorem:
Corollary 42.4.3 (Clairaut’s theorem: mixed partials are symmetric) Let f : U → ℝ with
U ⊆ V open be twice differentiable. Then for any point p such that the quantities are
defined,
42.5Towards differential forms
This concludes the exposition of what the derivative really is: the key idea I want
to communicate in this chapter is that Df should be thought of as a map from
U → V∨.
The next natural thing to do is talk about integration. The correct way to do this is through
a so-called differential form: you’ll finally know what all those stupid dx’s and dy’s really
mean. (They weren’t just there for decoration!)
42.6A few harder problems to think about
Problem 42A⋆ (Chain rule).Let U1U2U3 be differentiable maps between open
sets of normed vector spaces Vi, and let h = g ∘ f. Prove the Chain Rule: for any point
p ∈ U1, we have
Problem 42B.Let U ⊆ V be open, and f : U → ℝ be differentiable k times. Show
that (Dkf)p is symmetric in its k arguments, meaning for any v1,…,vk∈ V and any
permutation σ on we have
43Differential forms
In this chapter, all vector spaces are finite-dimensional real inner product spaces. We first start
by (non-rigorously) drawing pictures of all the things that we will define in this chapter. Then
we re-do everything again in its proper algebraic context.
43.1Pictures of differential forms
Before defining a differential form, we first draw some pictures. The key thing to keep in mind
is
“The definition of a differential form is: something you can integrate.” — Joe Harris
We’ll assume that all functions are smooth, i.e. infinitely differentiable.
Let U ⊆ V be an open set of a vector space V . Suppose that we have a function f : U → ℝ,
i.e. we assign a value to every point of U.
Definition 43.1.1.A 0-formf on U is just a smooth function f : U → ℝ.
Thus, if we specify a finite set S of points in U we can “integrate” over S by just adding up
the values of the points:
So, a 0-form f lets us integrate over 0-dimensional “cells”.
But this is quite boring, because as we know we like to integrate over things like curves, not
single points. So, by analogy, we want a 1-form to let us integrate over 1-dimensional cells:
i.e. over curves. What information would we need to do that? To answer this, let’s
draw a picture of a curve c, which can be thought of as a function c : [0,1] → U.
We might think that we could get away with just specifying a number on every point of U
(i.e. a 0-form f), and then somehow “add up” all the values of f along the curve.
We’ll use this idea in a moment, but we can in fact do something more general.
Notice how when we walk along a smooth curve, at every point p we also have some
extra information: a tangent vector v. So, we can define a 1-form α as follows. A
0-form just took a point and gave a real number, but a 1-form will take both apoint and a tangent vector at that point, and spit out a real number. So a
1-form α is a smooth function on pairs (p,v), where v is a tangent vector at p, to ℝ.
Hence
Actually, for any point p, we will require that α(p,−) is a linear function in terms of the
vectors: i.e. we want for example that α(p,2v) = 2α(p,v). So it is more customary to think of
α as:
Definition 43.1.2.A 1-formα is a smooth function
Like with Df, we’ll use αp instead of α(p). So, at every point p, αp is some linear functional
that eats tangent vectors at p, and spits out a real number. Thus, we think of αp as an element
of V∨;
Next, we draw pictures of 2-forms. This should, for example, let us integrate over a blob (a
so-called 2-cell) of the form
i.e. for example, a square in U. In the previous example with 1-forms, we looked at tangent
vectors to the curve c. This time, at points we will look at pairs of tangent vectors in U: in the
same sense that lots of tangent vectors approximate the entire curve, lots of tiny squares will
approximate the big square in U.
So what should a 2-form β be? As before, it should start by taking a point p ∈ U, so βp is now a
linear functional: but this time, it should be a linear map on two vectors v and w. Here v and
w are not tangent so much as their span cuts out a small parallelogram. So, the right thing to
do is in fact consider
That is, to use the wedge product to get a handle on the idea that v and w span a
parallelogram. Another valid choice would have been (V ∧ V )∨; in fact, the two are
isomorphic, but it will be more convenient to write it in the former.
43.2Pictures of exterior derivatives
Next question:
How can we build a 1-form from a 0-form?
Let f be a 0-form on U; thus, we have a function f : U → ℝ. Then in fact there is a very
natural 1-form on U arising from f, appropriately called df. Namely, given a point p and a
tangent vector v, the differential form (df)p returns the change in f along v. In other words,
it’s just the total derivative (Df)p(v).
Thus, df measures “the change in f”.
Now, even if I haven’t defined integration yet, given a curve c from a point a to b, what do
you think
should be equal to? Remember that df is the 1-form that measures “infinitesimal change in
f”. So if we add up all the change in f along a path from a to b, then the answer we get
should just be
This is the first case of something we call Stokes’ theorem.
Generalizing, how should we get from a 1-form to a 2-form? At each point p, the 2-form β
gives a βp which takes in a “parallelogram” and returns a real number. Now suppose we have a
1-form α. Then along each of the edges of a parallelogram, with an appropriate sign
convention the 1-form α gives us a real number. So, given a 1-form α, we define dα to be the
2-form that takes in a parallelogram spanned by v and w, and returns the measure of αalong the boundary.
Now, what happens if you integrate df along the entire square c? The right picture is that, if
we think of each little square as making up the big square, then the adjacent boundaries
cancel out, and all we are left is the main boundary. This is again just a case of the so-called
Stokes’ theorem.
Image from [?]
43.3Differential forms
Prototypical example for this section:Algebraically, something that looks like fe1∨∧e2∨ + …,and geometrically, see the previous section.
Let’s now get a handle on what dx means. Fix a real vector space V of dimension n, and let
e1, …, en be a standard basis. Let U be an open set.
Definition 43.3.1.We define a differential k-formα on U to be a smooth (infinitely
differentiable) map α : U → Λk(V∨). (Here Λk(V∨) is the wedge product.)
Like with Df, we’ll use αp instead of α(p).
Example 43.3.2 (k-forms for k = 0,1)
(a)
A 0-form is just a function U → ℝ.
(b)
A 1-form is a function U → V∨. For example, the total derivative Df of a function
V → ℝ is a 1-form.
(c)
Let V = ℝ3 with standard basis e1, e2, e3. Then a typical 2-form is given by
where f,g,h : V → ℝ are smooth functions.
Now, by the projection principle (?? ) we only have to specify a function on each of
basis elements of Λk(V∨). So, take any basis {ei} of V , and take the usual basis for Λk(V∨) of
elements
Thus, a general k-form takes the shape
Since this is a huge nuisance to write, we will abbreviate this to just
where we understand the sum runs over I = (i1,…,ik), and deI represents ei1∨∧∧eik∨.
Now that we have an element Λk(V∨), what can it do? Well, first let me get the definition
on the table, then tell you what it’s doing.
Definition 43.3.3.For linear functions ξ1,…,ξk∈ V∨ and vectors v1,…,vk∈ V , set
You can check that this is well-defined under e.g. v ∧ w = −w ∧ v and so on.
Example 43.3.4 (Evaluation of a differential form) Set V = ℝ3. Suppose that at some
point p, the 2-form α returns
Let v1 = 3e1 + e2 + 4e3 and v2 = 8e1 + 9e2 + 5e3. Then
What does this definition mean? One way to say it is that
If I walk to a point p ∈ U, a k-form α will take in k vectors v1,…,vkand spit out anumber, which is to be interpreted as a (signed) volume.
Picture:
In other words, at every point p, we get a function αp. Then I can feed in k vectors to
αp and get a number, which I interpret as a signed volume of the parallelpiped
spanned by the {vi}’s in some way (e.g. the flux of a force field). That’s why αp as a
“function” is contrived to lie in the wedge product: this ensures that the notion of
“volume” makes sense, so that for example, the equality αp(v1,v2) = −αp(v2,v1)
holds.
This is what makes differential forms so fit for integration.
43.4Exterior derivatives
Prototypical example for this section:Possibly dx1 = e1∨.
We now define the exterior derivative df that we gave pictures of at the beginning of the
section. It turns out that the exterior derivative is easy to compute given explicit coordinates
to work with.
First, given a function f : U → ℝ, we define
In particular, suppose V = ℝn and f(x1,…,xn) = x1 (i.e. f = e1∨). Then:
Question 43.4.1.Show that for any p ∈ U,
Abuse of Notation 43.4.2.Unfortunately, someone somewhere decided it would be a
good idea to use “x1” to denote e1∨ (because obviously1x1 means “the function that takes (x1,…,xn) ∈ ℝn to x1”) and then decided that
This notation is so entrenched that I have no choice but to grudgingly accept it. Note
that it’s not even right, since technically it’s (dx1)p = e1∨; dx1 is a 1-form.
Remark 43.4.3 — This is the reason why we use the notation in calculus now: given,
say, f : ℝ → ℝ by f(x) = x2, it is indeed true that
and so by (more) abuse of notation we write df∕dx = 2x.
More generally, we can define the exterior derivative in terms of our basis e1, …, en as
follows: if α = ∑IfIdeI then we set
This doesn’t depend on the choice of basis.
Example 43.4.4 (Computing some exterior derivatives) Let V = ℝ3 with standard
basis e1, e2, e3. Let f(x,y,z) = x4 + y3 + 2xz. Then we compute
So surprisingly, d2f is the zero map. Here, we have exploited ?? for the first time, in writing
dx, dy, dz.
And in fact, this is always true in general:
Theorem 43.4.5 (Exterior derivative vanishes) Let α be any k-form. Then d2(α) = 0.
Even more succinctly,
The proof is left as ?? .
Exercise 43.4.6.Compare the statement d2= 0 to the geometric picture of a 2-form given atthe beginning of this chapter. Why does this intuitively make sense?
Here are some other properties of d:
As we just saw, d2 = 0.
For a k-form α and ℓ-form β, one can show that
If f : U → ℝ is smooth, then df = Df.
In fact, one can show that df as defined above is the unique map sending k-forms to
(k + 1)-forms with these properties. So, one way to define df is to take as axioms the bulleted
properties above and then declare d to be the unique solution to this functional
equation. In any case, this tells us that our definition of d does not depend on the basis
chosen.
Recall that df measures the change in boundary. In that sense, d2 = 0 is saying something
like “the boundary of the boundary is empty”. We’ll make this precise when we see Stokes’
theorem in the next chapter.
43.5Closed and exact forms
Let α be a k-form.
Definition 43.5.1.We say α is closed if dα = 0.
Definition 43.5.2.We say α is exact if for some (k − 1)-form β, dβ = α. If k = 0, α
is exact only when α = 0.
Question 43.5.3.Show that exact forms are closed.
A natural question arises: are there closed forms which are not exact? Surprisingly, the
answer to this question is tied to topology. Here is one important example.
Example 43.5.4 (The angle form) Let U = ℝ2∖{0}, and let 𝜃(p) be the angle formed
by the x-axis and the line from the origin to p.
The 1-form α : U → (ℝ2)∨ defined by
is called the angle form: given p ∈ U it measures the change in angle 𝜃(p) along a
tangent vector. So intuitively, “α = d𝜃”. Indeed, one can check directly that the angle
form is closed.
However, α is not exact: there is no global smooth function 𝜃 : U → ℝ having α as a
derivative. This reflects the fact that one can actually perform a full 2π rotation around
the origin, i.e. 𝜃 only makes sense mod 2π. Thus existence of the angle form α reflects
the possibility of “winding” around the origin.
So the key idea is that the failure of a closed form to be exact corresponds quite well with
“holes” in the space: the same information that homotopy and homology groups are
trying to capture. To draw another analogy, in complex analysis Cauchy-Goursat
only works when U is simply connected. The “hole” in U is being detected by the
existence of a form α. The so-called de Rham cohomology will make this relation
explicit.
43.6A few harder problems to think about
Problem 43A.Show directly that the angle form
is closed.
Problem 43B. Establish ?? , which states that d2 = 0.
44Integrating differential forms
We now show how to integrate differential forms over cells, and state Stokes’ theorem in this
context. In this chapter, all vector spaces are finite-dimensional and real.
44.1Motivation: line integrals
Given a function g : [a,b] → ℝ, we know by the fundamental theorem of calculus
that
where f is a function such that g = df∕dt. Equivalently, for f : [a,b] → ℝ,
where df is the exterior derivative we defined earlier.
Cool, so we can integrate over [a,b]. Now suppose more generally, we have U an open subset
of our real vector space V and a 1-form α : U → V∨. We consider a parametrized curve,
which is a smooth function c : [a,b] → U. Picture:
We want to define an ∫cα such that:
The integral∫cα should add up all the α along the curve c.
Our differential form α first takes in a point p to get αp∈ V∨. Then, it eats a
tangent vector v ∈ V to the curve c to finally give a real number αp(v) ∈ ℝ. We
would like to “add all these numbers up”, using only the notion of an integral over
[a,b].
Exercise 44.1.1.Try to guess what the definition of the integral should be. (By type-checking,there’s only one reasonable answer.)
So, the definition we give is
Here, c′(t) is shorthand for (Dc)c(t)(1). It represents the tangent vector to the curve c at the
point p = c(t), at time t. (Here we are taking advantage of the fact that [a,b] is
one-dimensional.)
Now that definition was a pain to write, so we will define a differential 1-form
c∗α on [a,b] to swallow that entire thing: specifically, in this case we define c∗α to
be
(here 𝜀 is some displacement in time). Thus, we can more succinctly write
This is a special case of a pullback: roughly, if ϕ : U → U′ (where U ⊆ V , U′⊆ V ′),
we can change any differential k-form α on U′ to a k-form on U. In particular, if
U = [a,b],1
we can resort to our old definition of an integral. Let’s now do this in full generality.
44.2Pullbacks
Let V and V ′ be finite dimensional real vector spaces (possibly different dimensions) and
suppose U and U′ are open subsets of each; next, consider a k-form α on U′.
Given a map ϕ : U → U′ we now want to define a pullback in much the same way as before.
Picture:
Well, there’s a total of about one thing we can do. Specifically: α accepts a point in U′ and
k tangent vectors in V ′, and returns a real number. We want ϕ∗α to accept a point in p ∈ U
and k tangent vectors v1,…,vk in V , and feed the corresponding information to
α.
Clearly we give the point q = ϕ(p). As for the tangent vectors, since we are interested in
volume, we take the derivative of ϕ at p, (Dϕ)p, which will scale each of our vectors vi into
some vector in the target V ′. To cut a long story short:
Definition 44.2.1.Given ϕ : U → U′ and α a k-form, we define the pullback
There is a more concrete way to define the pullback using bases. Suppose w1,…,wn is a basis
of V ′ and e1,…,em is a basis of V . Thus, by the projection principle (?? ) the map ϕ : V → V ′
can be thought of as
where each ϕi takes in a v ∈ V and returns a real number. We know also that α can be
written concretely as
Then, we define
A diligent reader can check these definitions are equivalent.
Example 44.2.2 (Computation of a pullback) Let V = ℝ2 with basis e1 and e2, and
suppose ϕ : V → V ′ is given by sending
where w1, w2, w3 is a basis for V ′. Consider the form αq = f(q)w1∧w3, where f : V ′→ ℝ.
Then
It turns out that the pullback basically behaves nicely as possible, e.g.
ϕ∗(cα + β) = cϕ∗α + ϕ∗β (linearity)
ϕ∗(α ∧ β) = (ϕ∗α) ∧ (ϕ∗β)
ϕ1∗(ϕ2∗(α)) = (ϕ2∘ ϕ1)∗(α) (naturality)
but I won’t take the time to check these here (one can verify them all by expanding with a
basis).
44.3Cells
Prototypical example for this section:A disk in ℝ2can be thought of as the cell
[0,R] × [0,2π] → ℝ2by (r,𝜃)(r cos𝜃)e1 + (r sin𝜃)e2.
Now that we have the notion of a pullback, we can define the notion of an integral for
more general spaces. Specifically, to generalize the notion of integrals we had before:
Definition 44.3.1.A k-cell is a smooth function c : [a1,b1] × [a2,b2] ×…[ak,bk] → V .
Example 44.3.2 (Examples of cells) Let V = ℝ2 for convenience.
(a)
A 0-cell consists of a single point.
(b)
As we saw, a 1-cell is an arbitrary curve.
(c)
A 2-cell corresponds to a 2-dimensional surface. For example, the map c : [0,R] ×
[0,2π] → V by
can be thought of as a disk of radius R.
Then, to define an integral
for a differential k-form α and a k-cell c : [0,1]k→ V , we simply take the pullback
Since c∗α is a k-form on the k-dimensional unit box, it can be written as
f(x1,…,xn) dx1∧∧ dxn, so the above integral can be written as
Example 44.3.3 (Area of a circle) Consider V = ℝ2 and let c : (r,𝜃)(r cos𝜃)e1 + (r sin𝜃)e2
on [0,R] × [0,2π] as before. Take the 2-form α which gives αp = e1∨∧e2∨ at every point p.
Then
c∗α
= ∧
= r(cos2𝜃 + sin2𝜃)(dr ∧ d𝜃)
= r dr ∧ d𝜃
Thus,
which is the area of a circle.
Here’s some geometric intuition for what’s happening. Given a k-cell in V , a differential
k-form α accepts a point p and some tangent vectors v1, …, vk and spits out a number
αp(v1,…,vk), which as before we view as a signed hypervolume. Then the integral adds up allthese infinitesimals across the entire cell. In particular, if V = ℝk and we take the form
α : pe1∨∧∧ek∨, then what these α’s give is the kth hypervolume of the cell. For this
reason, this α is called the volume form on ℝk.
You’ll notice I’m starting to play loose with the term “cell”: while the cell
c : [0,R] × [0,2π] → ℝ2 is supposed to be a function I have been telling you to think of it as a
unit disk (i.e. in terms of its image). In the same vein, a curve [0,1] → V should be thought of
as a curve in space, rather than a function on time.
This error turns out to be benign. Let α be a k-form on U and c : [a1,b1] ×× [ak,bk] → U a
k-cell. Suppose ϕ : [a1′,b1′] ×…[ak′,bk′] → [a1,b1] ×× [ak,bk]; it is a reparametrization if
ϕ is bijective and (Dϕ)p is always invertible (think “change of variables”); thus
is a k-cell as well. Then it is said to preserve orientation if det(Dϕ)p> 0 for all p and
reverse orientation if det(Dϕ)p< 0 for all p.
Exercise 44.3.4.Why is it that exactly one of these cases must occur?
Theorem 44.3.5 (Changing variables doesn’t affect integrals) Let c be a k-cell, α a k-form,
and ϕ a reparametrization. Then
Proof.Use naturality of the pullback to reduce it to the corresponding theorem in normal
calculus. □
So for example, if we had parametrized the unit circle as [0,1] × [0,1] → ℝ2 by
(r,t)R cos(2πt)e1 + R sin(2πt)e2, we would have arrived at the same result. So we really can
think of a k-cell just in terms of the points it specifies.
44.4Boundaries
Prototypical example for this section:The boundary of [a,b] is {b}−{a}. The boundary of asquare goes around its edge counterclockwise.
First, I introduce a technical term that lets us consider multiple cells at once.
Definition 44.4.1.A k-chainU is a formal linear combination of k-cells over U, i.e. a
sum of the form
where each ai∈ ℝ and ci is a k-cell. We define ∫cα = ∑iai∫ci.
In particular, a 0-chain consists of several points, each with a given weight.
Now, how do we define the boundary? For a 1-cell [a,b] → U, as I hinted earlier we
want the answer to be the 0-chain {c(b)}−{c(a)}. Here’s how we do it in general.
Definition 44.4.2.Suppose c : [0,1]k→ U is a k-cell. Then the boundary of c,
denoted ∂c : [0,1]k−1→ U, is the (k − 1)-chain defined as follows. For each i = 1,…,k
define
cistart(t1,…,tk−1)
= (t1,…,ti−1,0,ti,…,tk)
cistop(t1,…,tk−1)
= (t1,…,ti−1,1,ti,…,tk).
Then
Finally, the boundary of a chain is the sum of the boundaries of each cell (with the
appropriate weights). That is, ∂(∑aici) = ∑ai∂ci.
Question 44.4.3.Satisfy yourself that one can extend this definition to a k-cell c defined onc : [a1,b1] ×× [ak,bk] → V (rather than from [0,1]k→ V ).
Example 44.4.4 (Examples of boundaries) Consider the 2-cell c : [0,1]2→ ℝ2 shown below.
Here p1, p2, p3, p4 are the images of (0,0), (0,1), (1,0), (1,1), respectively. Then we can think
of ∂c as
where each “interval” represents the 1-cell shown by the reddish arrows on the right. We can
take the boundary of this as well, and obtain an empty chain as
Example 44.4.5 (Boundary of a unit disk) Consider the unit disk given by
The four parts of the boundary are shown in the picture below:
Note that two of the arrows more or less cancel each other out when they are integrated.
Moreover, we interestingly have a degenerate 1-cell at the center of the circle; it is a constant
function [0,1] → ℝ2 which always gives the origin.
Obligatory theorem, analogous to d2 = 0 and left as a problem.
Theorem 44.4.6 (The boundary of the boundary is empty) ∂2 = 0, in the sense that for
any k-chain c we have ∂2(c) = 0.
44.5Stokes’ theorem
Prototypical example for this section:∫[a,b]dg = g(b) − g(a).
We now have all the ingredients to state Stokes’ theorem for cells.
Theorem 44.5.1 (Stokes’ theorem for cells) Take U ⊆ V as usual, let c : [0,1]k→ U be a
k-cell and let α : U → Λk−1(V∨) be a k − 1-form. Then
In particular, if dα = 0 then the left-hand side vanishes.
For example, if c is the interval [a,b] then ∂c = {b}−{a}, and thus we obtain the
fundamental theorem of calculus.
44.6A few harder problems to think about
Problem 44A† (Green’s theorem).Let f,g : ℝ2→ ℝ be smooth functions. Prove that
Problem 44B.Show that ∂2 = 0.
Problem 44C (Pullback and d commute).Let U and U′ be open sets of vector spaces
V and V ′ and let ϕ : U → U′ be a smooth map between them. Prove that for any
differential form α on U′ we have
Problem 44D (Arc length isn’t a form).Show that there does not exist a 1-form α on ℝ2
such that for a curve c : [0,1] → ℝ2, the integral ∫cα gives the arc length of c.
Problem 44E.An exactk-form α is one satisfying α = dβ for some β. Prove that
where C1 and C2 are any concentric circles in the plane and α is some exact 1-form.
45A bit of manifolds
Last chapter, we stated Stokes’ theorem for cells. It turns out there is a much larger class of
spaces, the so-called smooth manifolds, for which this makes sense.
Unfortunately, the definition of a smooth manifold is complete garbage, and so by the
time I am done defining differential forms and orientations, I will be too lazy to
actually define what the integral on it is, and just wave my hands and state Stokes’
theorem.
45.1Topological manifolds
Prototypical example for this section:S2: “the Earth looks flat”.
Long ago, people thought the Earth was flat, i.e. homeomorphic to a plane, and in
particular they thought that π2(Earth) = 0. But in fact, as most of us know, the Earth is
actually a sphere, which is not contractible and in particular π2(Earth)ℤ. This observation
underlies the definition of a manifold:
An n-manifold is a space which locally looks like ℝn.
Actually there are two ways to think about a topological manifold M:
“Locally”: at every point p ∈ M, some open neighborhood of p looks like an open
set of ℝn. For example, to someone standing on the surface of the Earth, the Earth
looks much like ℝ2.
“Globally”: there exists an open cover of M by open sets {Ui}i (possibly infinite)
such that each Ui is homeomorphic to some open subset of ℝn. For example, from
outer space, the Earth can be covered by two hemispherical pancakes.
Question 45.1.1.Check that these are equivalent.
While the first one is the best motivation for examples, the second one is easier to use
formally.
Definition 45.1.2.A topological n-manifoldM is a Hausdorff space with an open
cover {Ui} of sets homeomorphic to subsets of ℝn, say by homeomorphisms
where each Ei is an open subset of ℝn. Each ϕi : Ui→ Ei is called a chart, and
together they form a so-called atlas.
Remark 45.1.3 — Here “E” stands for “Euclidean”. I think this notation is not
standard; usually people just write ϕi(Ui) instead.
Remark 45.1.4 — This definition is nice because it doesn’t depend on embeddings: a
manifold is an intrinsic space M, rather than a subset of ℝN for some N. Analogy: an
abstract group G is an intrinsic object rather than a subgroup of Sn.
Example 45.1.5 (An atlas on S1) Here is a picture of an atlas for S1, with two open sets.
Question 45.1.6.Where do you think the words “chart” and “atlas” come from?
Example 45.1.7 (Some examples of topological manifolds)
(a)
As discussed at length, the sphere S2 is a 2-manifold: every point in the sphere has
a small open neighborhood that looks like D2. One can cover the Earth with just
two hemispheres, and each hemisphere is homeomorphic to a disk.
(b)
The circle S1 is a 1-manifold; every point has an open neighborhood that looks like
an open interval.
(c)
The torus, Klein bottle, ℝℙ2 are all 2-manifolds.
(d)
ℝn is trivially a manifold, as are its open sets.
All these spaces are compact except ℝn.
A non-example of a manifold is Dn, because it has a boundary; points on the boundary do not
have open neighborhoods that look Euclidean.
45.2Smooth manifolds
Prototypical example for this section:All the topological manifolds.
Let M be a topological n-manifold with atlas {UiEi}.
Definition 45.2.1.For any i, j such that Ui∩ Uj≠∅, the transition mapϕij is the
composed map
Sorry for the dense notation, let me explain. The intersection with the image ϕiimg(Ui∩Uj)
and the image ϕjimg(Ui∩Uj) is a notational annoyance to make the map well-defined and a
homeomorphism. The transition map is just the natural way to go from Ei→ Ej, restricted to
overlaps. Picture below, where the intersections are just the green portions of each E1 and
E2:
We want to add enough structure so that we can use differential forms.
Definition 45.2.2.We say M is a smooth manifold if all its transition maps are
smooth.
This definition makes sense, because we know what it means for a map between two open
sets of ℝn to be differentiable.
With smooth manifolds we can try to port over definitions that we built for ℝn onto our
manifolds. So in general, all definitions involving smooth manifolds will reduce to something
on each of the coordinate charts, with a compatibility condition.
AS an example, here is the definition of a “smooth map”:
Definition 45.2.3.
(a)
Let M be a smooth manifold. A continuous function f : M → ℝ is called smooth
if the composition
is smooth as a function Ei→ ℝ.
(b)
Let M and N be smooth with atlases {UiMEiM}i and {UjNEiN}j, A map
f : M → N is smooth if for every i and j, the composed map
is smooth, as a function Ei→ Ej.
45.3Regular value theorem
Prototypical example for this section:x2 + y2 = 1 is a circle!
Despite all that I’ve written about general manifolds, it would be sort of mean if I left you
here because I have not really told you how to actually construct manifolds in practice, even
though we know the circle x2 + y2 = 1 is a great example of a one-dimensional manifold
embedded in ℝ2.
Theorem 45.3.1 (Regular value theorem) Let V be an n-dimensional real normed vector
space, let U ⊆ V be open and let f1,…,fm: U → ℝ be smooth functions. Let M be the
set of points p ∈ U such that f1(p) = = fm(p) = 0.
Assume M is nonempty and that the map
has rank m, for every point p ∈ M. Then M is a manifold of dimension n − m.
For a proof, see [?, Theorem 6.3].
One very common special case is to take m = 1 above.
Corollary 45.3.2 (Level hypersurfaces) Let V be a finite-dimensional real normed vector
space, let U ⊆ V be open and let f : U → ℝ be smooth. Let M be the set of points
p ∈ U such that f(p) = 0. If M≠∅ and (Df)p is not the zero map for any p ∈ M, then
M is a manifold of dimension n − 1.
Example 45.3.3 (The circle x2 + y2− c = 0) Let f(x,y) = x2 + y2− c, f : ℝ2→ ℝ,
where c is a positive real number. Note that
which in particular is nonzero as long as (x,y)≠(0,0), i.e. as long as c≠0. Thus:
When c > 0, the resulting curve — a circle with radius — is a one-dimensional
manifold, as we knew.
When c = 0, the result fails. Indeed, M is a single point, which is actually a
zero-dimensional manifold!
We won’t give further examples since I’m only mentioning this in passing in order to
increase your capacity to write real concrete examples. (But [?, Chapter 6.2] has some more
examples, beautifully illustrated.)
45.4Differential forms on manifolds
We already know what a differential form is on an open set U ⊆ ℝn. So, we naturally try to
port over the definition of differentiable form on each subset, plus a compatibility
condition.
Let M be a smooth manifold with atlas {UiEi}i.
Definition 45.4.1.A differential k-formα on a smooth manifold M is a collection
{αi}i of differential k-forms on each Ei, such that for any j and i we have that
In English: we specify a differential form on each chart, which is compatible under pullbacks of
the transition maps.
45.5Orientations
Prototypical example for this section:Left versus right, clockwise vs. counterclockwise.
This still isn’t enough to integrate on manifolds. We need one more definition: that of an
orientation.
The main issue is the observation from standard calculus that
Consider then a space M which is homeomorphic to an interval. If we have a 1-form α, how
do we integrate it over M? Since M is just a topological space (rather than a subset of ℝ),
there is no default “left” or “right” that we can pick. As another example, if M = S1 is a
circle, there is no default “clockwise” or “counterclockwise” unless we decide to embed M into
ℝ2.
To work around this we have to actually have to make additional assumptions about our
manifold.
Definition 45.5.1.A smooth n-manifold is orientable if there exists a differential
n-form ω on M such that for every p ∈ M,
Recall here that ωp is an element of Λn(V∨). In that case we say ω is a volume form of
M.
How do we picture this definition? If we recall that an differential form is supposed to take
tangent vectors of M and return real numbers. To this end, we can think of each point p ∈ M
as having a tangent planeTp(M) which is n-dimensional. Now since the volume form ω is
n-dimensional, it takes an entire basis of the Tp(M) and gives a real number. So a manifold is
orientable if there exists a consistent choice of sign for the basis of tangent vectors at every
point of the manifold.
For “embedded manifolds”, this just amounts to being able to pick a nonzero field
of normal vectors to each point p ∈ M. For example, S1 is orientable in this way.
Similarly, one can orient a sphere S2 by having a field of vectors pointing away (or towards) the
center. This is all non-rigorous, because I haven’t defined the tangent plane Tp(M); since M is
in general an intrinsic object one has to be quite roundabout to define Tp(M) (although I do
so in an optional section later). In any event, the point is that guesses about the orientability
of spaces are likely to be correct.
Example 45.5.2 (Orientable surfaces)
(a)
Spheres Sn, planes, and the torus S1× S1 are orientable.
(b)
The Möbius strip and Klein bottle are not orientable: they are “one-sided”.
(c)
ℂℙn is orientable for any n.
(d)
ℝℙn is orientable only for odd n.
45.6Stokes’ theorem for manifolds
Stokes’ theorem in the general case is based on the idea of a manifold with boundaryM,
which I won’t define, other than to say its boundary ∂M is an n− 1 dimensional manifold, and
that it is oriented if M is oriented. An example is M = D2, which has boundary
∂M = S1.
Next,
Definition 45.6.1.The support of a differential form α on M is the closure of the set
If this support is compact as a topological space, we say α is compactly supported.
Remark 45.6.2 — For example, volume forms are supported on all of M.
Now, one can define integration on oriented manifolds, but I won’t define this because the
definition is truly awful. Then Stokes’ theorem says
Theorem 45.6.3 (Stokes’ theorem for manifolds) Let M be a smooth oriented n-manifold
with boundary and let α be a compactly supported n − 1-form. Then
All the omitted details are developed in full in [?].
45.7(Optional) The tangent and contangent space
Prototypical example for this section:Draw a line tangent to a circle, or a plane tangent to asphere.
Let M be a smooth manifold and p ∈ M a point. I omitted the definition of Tp(M) earlier,
but want to actually define it now.
As I said, geometrically we know what this should look like for our usual examples. For
example, if M = S1 is a circle embedded in ℝ2, then the tangent vector at a point p should
just look like a vector running off tangent to the circle. Similarly, given a sphere M = S2, the
tangent space at a point p along the sphere would look like plane tangent to M at
p.
However, one of the points of all this manifold stuff is that we really want to see
the manifold as an intrinsic object, in its own right, rather than as embedded in
ℝn.1
So, we would like our notion of a tangent vector to not refer to an ambient space, but only to
intrinsic properties of the manifold M in question.
45.7.iTangent space
To motivate this construction, let us start with an embedded case for which we know the
answer already: a sphere.
Suppose f : S2→ ℝ is a function on a sphere, and take a point p. Near the point
p, f looks like a function on some open neighborhood of the origin. Thus we can
think of taking a directional derivative along a vector v in the imagined tangent
plane (i.e. some partial derivative). For a fixed v this partial derivative is a linear
map
It turns out this goes the other way: if you know what Dv does to every smooth function,
then you can recover v. This is the trick we use in order to create the tangent space. Rather
than trying to specify a vector v directly (which we can’t do because we don’t have an ambient
space),
The vectors are partial-derivative-like maps.
More formally, we have the following.
Definition 45.7.1.A derivationD at p is a linear map D: C∞(M) → ℝ
(i.e. assigning a real number to every smooth f) satisfying the following Leibniz rule:
for any f, g we have the equality
This is just a “product rule”. Then the tangent space is easy to define:
Definition 45.7.2.A tangent vector is just a derivation at p, and the tangent spaceTp(M) is simply the set of all these tangent vectors.
In this way we have constructed the tangent space.
45.7.iiThe cotangent space
In fact, one can show that the product rule for D is equivalent to the following three
conditions:
1.
D is linear, meaning D(af + bg) = aD(f) + bD(g).
2.
D(1M) = 0, where 1M is the constant function on M.
3.
D(fg) = 0 whenever f(p) = g(p) = 0. Intuitively, this means that if a function
h = fg vanishes to second order at p, then its derivative along D should be zero.
This suggests a third equivalent definition: suppose we define
to be the set of functions which vanish at p (this is called the maximal ideal at p). In that
case,
is the set of functions vanishing to second order at p. Thus, a tangent vector is really just a
linear map
In other words, the tangent space is actually the dual space of 𝔪p∕𝔪p2; for this reason, the
space 𝔪p∕𝔪p2 is defined as the cotangent space (the dual of the tangent space). This
definition is even more abstract than the one with derivations above, but has some nice
properties:
it is coordinate-free, and
it’s defined only in terms of the smooth functions M → ℝ, which will be really
helpful later on in algebraic geometry when we have varieties or schemes and can
repeat this definition.
45.7.iiiSanity check
With all these equivalent definitions, the last thing I should do is check that this definition of
tangent space actually gives a vector space of dimension n. To do this it suffices to show verify
this for open subsets of ℝn, which will imply the result for general manifolds M (which are
locally open subsets of ℝn). Using some real analysis, one can prove the following
result:
Theorem 45.7.3 Suppose M ⊂ ℝn is open and 0 ∈ M. Then
In other words 𝔪02 is the set of functions which vanish at 0 and such that all first
derivatives of f vanish at zero.
Thus, it follows that there is an isomorphism
and so the cotangent space, hence tangent space, indeed has dimension n.
45.8A few harder problems to think about
Problem 45A.Show that a differential 0-form on a smooth manifold M is the same
thing as a smooth function M → ℝ.
Part XIII Algebraic NT I: Rings of Integers
46Algebraic integers
Here’s a first taste of algebraic number theory.
This is really close to the border between olympiads and higher math. You’ve always known
that a + b had a “norm” a2− 2b2, and that somehow this norm was multiplicative. You’ve
also always known that roots come in conjugate pairs. You might have heard of minimal
polynomials but not know much about them.
This chapter and the next one will make all these vague notions precise. It’s drawn largely
from the first chapter of [?].
46.1Motivation from high school algebra
This is adapted from my blog, PowerOverwhelming1 .
In high school precalculus, you’ll often be asked to find the roots of some polynomial with
integer coefficients. For instance,
has roots 3, −1 + 2i, −1 − 2i. Or as another example,
has roots −1, 2 + , 2 −. You’ll notice that the irrational roots, like −1 ± 2i and
2 ±, are coming up in pairs. In fact, I think precalculus explicitly tells you that the
imaginary roots come in conjugate pairs. More generally, it seems like all the roots of the form
a + b come in “conjugate pairs”. And you can see why.
But a polynomial like
has no rational roots. (The roots of this are approximately −3.0514, 0.51730, 2.5341.) Or
even simpler,
has only one real root, . These roots, even though they are irrational, have no
“conjugate” pairs. Or do they?
Let’s try and figure out exactly what’s happening. Let α be any complex number. We define
a minimal polynomial of α over ℚ to be a polynomial such that
P(x) has rational coefficients, and leading coefficient 1,
P(α) = 0.
The degree of P is as small as possible. We call deg P the degree of α.
Example 46.1.1 (Examples of minimal polynomials)
(a)
has minimal polynomial x2− 2.
(b)
The imaginary unit i = has minimal polynomial x2 + 1.
(c)
A primitive pth root of unity, ζp = e, has minimal polynomial xp−1+xp−2++1,
where p is a prime.
Note that 100x2− 200 is also a polynomial of the same degree which has as a root; that’s
why we want to require the polynomial to be monic. That’s also why we choose to work in the
rational numbers; that way, we can divide by leading coefficients without worrying if we get
non-integers.
Why do we care? The point is as follows: suppose we have another polynomial A(x) such
that A(α) = 0. Then we claim that P(x) actually divides A(x)! That means that all the other
roots of P will also be roots of A.
The proof is by contradiction: if not, by polynomial long division we can find a quotient and
remainder Q(x), R(x) such that
and R(x)≢0. Notice that by plugging in x = α, we find that R(α) = 0. But deg R < deg P,
and P(x) was supposed to be the minimal polynomial. That’s impossible!
It follows from this and the monotonicity of the minimal polynomial that it is unique (when
it exists), so actually it is better to refer to the minimal polynomial.
Exercise 46.1.2.Can you find an element in ℂ that has no minimal polynomial?
Let’s look at a more concrete example. Consider A(x) = x3− 3x2− 2x + 2 from the
beginning. The minimal polynomial of 2 + is P(x) = x2− 4x + 2 (why?). Now we know
that if 2 + is a root, then A(x) is divisible by P(x). And that’s how we know that if 2 +
is a root of A, then 2 − must be a root too.
As another example, the minimal polynomial of is x3− 2. So actually has two
conjugates, namely, α = and β = . Thus any
polynomial which vanishes at also has α and β as roots!
Question 46.1.3 (Important but tautological: irreducible⇔minimal).Let α be a root of thepolynomial P(x). Show that P(x) is the minimal polynomial if and only if it is irreducible.
46.2Algebraic numbers and algebraic integers
Prototypical example for this section:is an algebraic integer (root of x2− 2),is analgebraic number but not an algebraic integer (root of x −).
Let’s now work in much vaster generality. First, let’s give names to the new numbers we’ve
discussed above.
Definition 46.2.1.An algebraic number is any α ∈ ℂ which is the root of some
polynomial with coefficients in ℚ. The set of algebraic numbers is denoted ℚ.
Remark 46.2.2 — One can equally well say algebraic numbers are those that are roots
of some polynomial with coefficients in ℤ (rather than ℚ), since any polynomial in ℚ[x]
can be scaled to one in ℤ[x].
Definition 46.2.3.Consider an algebraic number α and its minimal polynomial P
(which is monic and has rational coefficients). If it turns out the coefficients of P are
integers, then we say α is an algebraic integer.
The set of algebraic integers is denoted ℤ.
Remark 46.2.4 — One can show, using Gauss’s Lemma, that if α is the root of any
monic polynomial with integer coefficients, then α is an algebraic integer. So in practice,
if I want to prove that + is an algebraic integer, then I only have to say “the
polynomial (x2− 5)2− 24 works” without checking that it’s minimal.
Sometimes for clarity, we refer to elements of ℤ as rational integers.
Example 46.2.5 (Examples of algebraic integers) The numbers
are all algebraic integers, since they are the roots of the monic polynomials x−4, x2 +1,
x3− 2 and (x2− 5)2− 24.
The number has minimal polynomial x −, so it’s an algebraic number but not an
algebraic integer. (In fact, the rational root theorem also directly implies that any monic
integer polynomial does not have as a root!)
There are two properties I want to give for these off the bat, because they’ll be used
extensively in the tricky (but nice) problems at the end of the section. The first we prove now,
since it’s very easy:
Proposition 46.2.6 (Rational algebraic integers are rational integers) An algebraic integer
is rational if and only if it is a rational integer. In symbols,
Proof.Let α be a rational number. If α is an integer, it is the root of x − α, hence an
algebraic integer too.
Conversely, if P is a monic polynomial with integer coefficients such that P(α) = 0
then (by the rational root theorem, say) it follows α must be an integer. □
The other is that:
Proposition 46.2.7 (ℤis a ring and ℚis a field) The algebraic integers ℤ form a ring. The
algebraic numbers ℚ form a field.
We could prove this now if we wanted to, but the results in the next chapter will more or
less do it for us, and so we take this on faith temporarily.
46.3Number fields
Prototypical example for this section:ℚ() is a typical number field.
Given any algebraic number α, we’re able to consider fields of the form ℚ(α). Let us write
down the more full version.
Definition 46.3.1.A number fieldK is a field containing ℚ as a subfield which is a
finite-dimensional ℚ-vector space. The degree of K is its dimension.
Example 46.3.2 (Prototypical example) Consider the field
This is a field extension of ℚ, and has degree 2 (the basis being 1 and ).
You might be confused that I wrote ℚ() (which should permit denominators) instead of
ℚ[], say. But if you read through ?? , you should see that the denominators don’t
really matter: = (3 + ) anyways, for example. You can either check this
now in general, or just ignore the distinction and pretend I wrote square brackets
everywhere.
Exercise 46.3.3 (Unimportant).Show that if α is an algebraic number, then ℚ(α)ℚ[α].
Example 46.3.4 (Adjoining an algebraic number) Let α be the root of some irreducible
polynomial P(x) in ℚ. The field ℚ(α) is a field extension as well, and the basis is
1,α,α2,…,αm−1, where m is the degree of α. In particular, the degree of ℚ(α) is just the
degree of α.
Example 46.3.5 (Non-examples of number fields) ℝ and ℂ are not number fields since
there is no finite ℚ-basis of them.
46.4Primitive element theorem, and monogenic extensions
Prototypical example for this section:ℚ(,)ℚ( + ). Can you see why?
I’m only putting this theorem here because I was upset that no one told me it was true (it’s
a very natural conjecture), and I hope to not do the same to the reader. However, I’m not
going to use it in anything that follows.
Theorem 46.4.1 (Artin’s primitive element theorem) Every number field K is isomorphic
to ℚ(α) for some algebraic number α.
The proof is left as ?? , since to prove it I need to talk about field extensions
first.
The prototypical example
makes it clear why this theorem should not be too surprising.
46.5A few harder problems to think about
Problem 46A.Find a polynomial with integer coefficients which has + as a
root.
Problem 46B (Brazil 2006).Let p be an irreducible polynomial in ℚ[x]
and degree larger than 1. Prove that if p has two roots r and s whose product is 1 then the
degree of p is even.
Problem 46C⋆. Consider n roots of unity 𝜀1, …, 𝜀n. Assume the average (𝜀1 + + 𝜀n) is
an algebraic integer. Prove that either the average is zero or 𝜀1 = = 𝜀n. (Used in ?? .)
Problem 46D†.Which rational numbers q satisfy cos(qπ) ∈ ℚ?
Problem 46E (MOP 2010).There are n > 2 lamps arranged in a circle; initially one is on
and the others are off. We may select any regular polygon whose vertices are among the lamps
and toggle the states of all the lamps simultaneously. Show it is impossible to turn all lamps
off.
Problem 46F (Kronecker’s theorem).Let α be an algebraic integer.
Suppose all its Galois conjugates have absolute value one. Prove that αN = 1 for some positive
integer N.
Problem 46G.Is there an algebraic integer with absolute value one
which is not a root of unity?
Problem 46H.Is the ring of algebraic integers Noetherian?
47The ring of integers
47.1Norms and traces
Prototypical example for this section:a + bas an element of ℚ() has norm a2− 2b2andtrace 2a.
Remember when you did olympiads and we had like a2 + b2 was the “norm” of a + bi? Cool,
let me tell you what’s actually happening.
First, let me make precise the notion of a conjugate.
Definition 47.1.1.Let α be an algebraic number, and let P(x) be its minimal
polynomial, of degree m. Then the m roots of P are the (Galois) conjugates of α.
It’s worth showing at the moment that there are no repeated conjugates.
Lemma 47.1.2 (Irreducible polynomials have distinct roots) An irreducible polynomial in
ℚ[x] cannot have a complex double root.
Proof.Let f(x) ∈ ℚ[x] be the irreducible polynomial and assume it has a double root
α. Take the derivative f′(x). This derivative has three interesting properties.
The degree of f′ is one less than the degree of f.
The polynomials f and f′ are not relatively prime because they share a factor
x − α.
The coefficients of f′ are also in ℚ.
Consider g = gcd(f,f′). We must have g ∈ ℚ[x] by Euclidean algorithm. But the first two facts
about f′ ensure that g is nonconstant and deg g < deg f. Yet g divides f, contradiction to the
fact that f should be a minimal polynomial. □
Hence α has exactly as many conjugates as the degree of α.
Now, we would like to define the norm of an element N(α) as the product of its conjugates.
For example, we want 2 + i to have norm (2 + i)(2 − i) = 5, and in general for a + bi to have
norm a2 + b2. It would be really cool if the norm was multiplicative; we already know this is
true for complex numbers!
Unfortunately, this doesn’t quite work: consider
But (2 + i)(2 −i) = 5, which doesn’t have norm 25 like we want, since 5 is degree 1 and has
no conjugates at all. The reason this “bad” thing is happening is that we’re trying to define
the norm of an element, when we really ought to be defining the norm of an element withrespect to a particular K.
What I’m driving at is that the norm should have different meanings depending on which
field you’re in. If we think of 5 as an element of ℚ, then its norm is 5. But thought of as an
element of ℚ(i), its norm really ought to be 25. Let’s make this happen: for K a number
field, we will now define NK∕ℚ(α) to be the norm of α with respect to K as follows.
Definition 47.1.3.Let α ∈ K have degree n, so ℚ(α) ⊆ K, and set k = (deg K)∕n.
The norm of α is defined as
The trace is defined as
The exponent of k is a “correction factor” that makes the norm of 5 into 52 = 25 when we
view 5 as an element of ℚ(i) rather than an element of ℚ. For a “generic” element of K, we
expect k = 1.
Exercise 47.1.4.Use what you know about nested vector spaces to convince yourself that k isactually an integer.
Example 47.1.5 (Norm of a + b) Let α = a+b∈ ℚ() = K. If b≠0, then α and
K have the degree 2. Thus the only conjugates of α are a ± b, which gives the norm
The trace is (a − b) + (a + b) = 2a.
Nicely, the formula a2− 2b2 and 2a also works when b = 0.
Of importance is:
Proposition 47.1.6 (Norms and traces are rational integers) If α is an algebraic integer, its
norm and trace are rational integers.
Question 47.1.7.Prove it. (Vieta formula.)
That’s great, but it leaves a question unanswered: why is the norm multiplicative? To do
this, I have to give a new definition of norm and trace.
Theorem 47.1.8 (Morally correct definition of norm and trace) Let K be a number field of
degree n, and let α ∈ K. Let μα: K → K denote the map
viewed as a linear map of ℚ-vector spaces. Then,
the norm of α equals the determinant detμα, and
the trace of α equals the trace Trμα.
Since the trace and determinant don’t depend on the choice of basis, you can pick
whatever basis you want and use whatever definition you got in high school. Fantastic,
right?
Example 47.1.9 (Explicit computation of matrices for a + b) Let K = ℚ(), and
let 1, be the basis of K. Let
(possibly even b = 0), and notice that
We can rewrite this in matrix form as
Consequently, we can interpret μα as the matrix
Of course, the matrix will change if we pick a different basis, but the determinant and
trace do not: they are always given by
This interpretation explains why the same formula should work for a + b even in the case
b = 0.
Proof.I’ll prove the result for just the norm; the trace falls out similarly. Set
The proof is split into two parts, depending on whether or not k = 1.
Proof if k = 1. Set n = deg α = deg K. Thus the norm actually is the product of the
Galois conjugates. Also,
is linearly independent in K, and hence a basis (as dimK = n). Let’s use this as the
basis for μα.
Let
be the minimal polynomial of α. Thus μα(1) = α, μα(α) = α2, and so on, but
μα(αn−1) = −cn−1αn−1−− c0. Therefore, μα is given by the matrix
Thus
and we’re done by Vieta’s formulas. ■
Proof if k > 1. We have nested vector spaces
Let e1, …, ek be a ℚ(α)-basis for K (meaning: interpret K as a vector space over ℚ(α),
and pick that basis). Since {1,α,…,αn−1} is a ℚ basis for ℚ(α), the elements
constitute a ℚ-basis of K. Using this basis, the map μα looks like
where M is the same matrix as above: we just end up with one copy of our old matrix
for each ei. Thus detμα = (detM)k, as needed. ■
Question 47.1.10.Verify the result for traces as well.□
From this it follows immediately that
because by definition we have
and that the determinant is multiplicative. In the same way, the trace is additive.
47.2The ring of integers
Prototypical example for this section:If K = ℚ(), then 𝒪K = ℤ[]. But if K = ℚ(),then 𝒪K = ℤ[].
ℤ makes for better number theory than ℚ. In the same way, focusing on the algebraicintegers of K gives us some really nice structure, and we’ll do that here.
Definition 47.2.1.Given a number field K, we define
to be the ring of integers of K; in other words 𝒪K consists of the algebraic integers
of K.
We do the classical example of a quadratic field now. Before proceeding, I need to write a
silly number theory fact.
Exercise 47.2.2 (Annoying but straightforward).Let a and b be rational numbers, and d a squarefreepositive integer.
If d ≡ 2,3 (mod4), prove that 2a,a2− db2∈ℤ if and only if a,b ∈ℤ.
For d ≡ 1 (mod4), prove that 2a,a2−db2∈ℤ if and only if a,b ∈ℤ OR if a−,b−∈ℤ.
You’ll need to take mod 4.
Example 47.2.3 (Ring of integers of K = ℚ()) Let K be as above. We claim that
We set α = a+b. Then α ∈𝒪K when the minimal polynomial has integer coefficients.
If b = 0, then the minimal polynomial is x − α = x − a, and thus α works if and only if
it’s an integer. If b≠0, then the minimal polynomial is
From the exercise, this occurs exactly for a,b ∈ ℤ.
Example 47.2.4 (Ring of integers of K = ℚ()) We claim that in this case
The proof is exactly the same, except the exercise tells us instead that for b≠0, we have
both the possibility that a,b ∈ ℤ or that a,b ∈ ℤ −. This reflects the fact that is
the root of x2− x − 1 = 0; no such thing is possible with .
In general, the ring of integers of K = ℚ() is
What we’re going to show is that 𝒪K behaves in K a lot like the integers do in
ℚ. First we show K consists of quotients of numbers in 𝒪K. In fact, we can do
better:
Example 47.2.5 (Rationalizing the denominator) For example, consider K = ℚ().
The number x = is an element of K, but by “rationalizing the denominator” we
can write
So we see that in fact, x is of an integer in 𝒪K.
The theorem holds true more generally.
Theorem 47.2.6 (K = ℚ ⋅𝒪K) Let K be a number field, and let x ∈ K be any element.
Then there exists an integer n such that nx ∈𝒪K; in other words,
for some α ∈𝒪K.
Exercise 47.2.7.Prove this yourself. (Start by using the fact that x has a minimal polynomialwith rational coefficients. Alternatively, take the norm.)
Now we are going to show 𝒪K is a ring; we’ll check it is closed under addition and
multiplication. To do so, the easiest route is:
Lemma 47.2.8 (α ∈ℤ⇔ℤ[α] finitely generated) Let α ∈ℚ. Then α is an algebraic
integer if and only if the abelian group ℤ[α] is finitely generated.
Proof.Note that α is an algebraic integer if and only if it’s the root of some nonzero,
monic polynomial with integer coefficients. Suppose first that
Then the set 1,α,…,αN−1 generates ℤ[α], since we can repeatedly replace αN until all
powers of α are less than N.
Conversely, suppose that ℤ[α] is finitely generated by some b1,…,bm. Viewing the bi as
polynomials in α, we can select a large integer N (say N = deg b1 ++deg bm +2015)
and express αN in the bi’s to get
The above gives us a monic polynomial in α, and the choice of N guarantees it is not
zero. So α is an algebraic integer. □
Example 47.2.9 ( isn’t an algebraic integer) We already know isn’t an algebraic
integer. So we expect
to not be finitely generated, and this is the case.
Question 47.2.10.To make the last example concrete: name all the elements of ℤ[] that cannotbe written as an integer combination of
Now we can state the theorem.
Theorem 47.2.11 (Algebraic integers are closed under + and ×) The set ℤ is closed under
addition and multiplication; i.e. it is a ring. In particular, 𝒪K is also a ring for any
number field K.
Proof.Let α,β ∈ℤ. Then ℤ[α] and ℤ[β] are finitely generated. Hence so is ℤ[α,β].
(Details: if ℤ[α] has ℤ-basis a1,…,am and ℤ[β] has ℤ-basis b1,…,bn, then take the mn
elements aibj.)
Now ℤ[α±β] and ℤ[αβ] are subsets of ℤ[α,β] and so they are also finitely generated.
Hence α ± β and αβ are algebraic integers. □
In fact, something even better is true. As you saw, for ℚ() we had 𝒪K = ℤ[]; in other
words, 𝒪K was generated by 1 and . Something similar was true for ℚ(). We claim that
in fact, the general picture looks exactly like this.
Theorem 47.2.12 (𝒪Kis a free ℤ-module of rank n) Let K be a number field of degree
n. Then 𝒪K is a free ℤ-module of rank n, i.e. 𝒪Kℤ⊕n as an abelian group. In other
words, 𝒪K has a ℤ-basis of n elements as
where αi are algebraic integers in 𝒪K.
Proof.TODO: add this in. (Originally, there was an incorrect proof; the mistake was
pointed out 2020-02-12 on https://math.stackexchange.com/q/3543641/229197 and
I hope to supply a correct one soon.) □
This last theorem shows that in many ways 𝒪K is a “lattice” in K. That is, for a number
field K we can find α1, …, αn in 𝒪K such that
𝒪K
α1ℤ ⊕ α2ℤ ⊕⊕ αnℤ
K
α1ℚ ⊕ α2ℚ ⊕⊕ αnℚ
as abelian groups.
47.3On monogenic extensions
Recall that it turned out number fields K could all be expressed as ℚ(α) for some α.
We might hope that something similar is true of the ring of integers: that we can
write
in which case {1,𝜃,…,𝜃n−1} serves both as a basis of K and as the ℤ-basis for 𝒪K (here
n = [K : ℚ]). In other words, we hope that the basis of 𝒪K is actually a “power
basis”.
This is true for the most common examples we use:
the quadratic field, and
the cyclotomic field in ?? .
Unfortunately, it is not true in general: the first counterexample is ℚ(α) for α a root of
X3− X2− 2X − 8.
We call an extension with this nice property monogenic. As we’ll later see, monogenic
extensions have a really nice factoring algorithm, ?? .
47.4A few harder problems to think about
Problem 47A⋆. Show that α is a unit of 𝒪K (meaning α−1∈𝒪K) if and only if
NK∕ℚ(α) = ±1.
Problem 47B⋆.Let K be a number field. What is the field of fractions of 𝒪K?
Problem 47C (Russian olympiad 1984).Find all integers m and n such that
Problem 47D (USA TST 2012).Decide whether there exist a,b,c > 2010 satisfying
Problem 47E† (Cyclotomic Field). Let p be an odd rational prime and ζp
a primitive pth root of unity. Let K = ℚ(ζp). Prove that 𝒪K = ℤ[ζp]. (In fact, the result is
true even if p is not a prime.)
48Unique factorization (finally!)
Took long enough.
48.1Motivation
Suppose we’re interested in solutions to the Diophantine equation n = x2 + 5y2 for a given n.
The idea is to try and “factor” n in ℤ[], for example
Unfortunately, this is not so simple, because as I’ve said before we don’t have unique
factorization of elements:
One reason this doesn’t work is that we don’t have a notion of a greatest common divisor.
We can write (35,77) = 7, but what do we make of (3,1 + )?
The trick is to use ideals as a “generalized GCD”. Recall that by (a,b) I mean the ideal
{ax + by∣x,y ∈ ℤ[]}. You can see that (35,77) = (7), but (3,1 + ) will be left
“unsimplified” because it doesn’t represent an actual value in the ring. Using these sets
(ideals) as elements, it turns out that we can develop a full theory of prime factorization, and
we do so in this chapter.
In other words, we use the ideal (a1,…,am) to interpret a “generalized GCD” of a1, …, am.
In particular, if we have a number x we want to represent, we encode it as just
(x).
Going back to our example of 6,
Please take my word for it that in fact, the complete prime factorization of (6) into prime
ideals is
In fact, (2) = 𝔭2, (3) = 𝔮1𝔮2, (1 + ) = 𝔭𝔮1, (1 −) = 𝔭𝔮2. So 6 indeed factorizes
uniquely into ideals, even though it doesn’t factor into elements.
As one can see above, ideal factorization is more refined than element factorization. Once
you have the factorization into ideals, you can from there recover all the factorizations into
elements. The upshot of this is that if we want to write n as x2 + 5y2, we just have to factor n
into ideals, and from there we can recover all factorizations into elements, and finally all ways
to write n as x2 + 5y2. Since we can already break n into rational prime factors (for example
6 = 2 ⋅ 3 above) we just have to figure out how each rational prime p∣n breaks down. There’s
a recipe for this, ?? ! In fact, I’ll even tell you what is says in this special case:
If t2 + 5 factors as (t + c)(t − c) (mod p), then (p) = (p,c + )(p,c −).
Otherwise, (p) is a prime ideal.
In this chapter we’ll develop this theory of unique factorization in full generality.
Remark 48.1.1 — In this chapter, I’ll be using the letters 𝔞, 𝔟, 𝔭, 𝔮 for ideals of 𝒪K.
When fractional ideals arise, I’ll use I and J for them.
48.2Ideal arithmetic
Prototypical example for this section: (x)(y) = (xy). In any case, think in terms ofgenerators.
First, I have to tell you how to add and multiply two ideals 𝔞 and 𝔟.
Definition 48.2.1.Given two ideals 𝔞 and 𝔟 of a ring R, we define
𝔞 + 𝔟
:=
𝔞 ⋅𝔟
: = .
(Note that infinite sums don’t make sense in general rings, which is why in 𝔞 ⋅𝔟 we cut off the
sum after some finite number of terms.) You can readily check these are actually ideals. This
definition is more natural if you think about it in terms of the generators of 𝔞 and
𝔟.
Proposition 48.2.2 (Ideal arithmetic via generators) Suppose 𝔞 = and
𝔟 = are ideals in a ring R. Then
(a)
𝔞 + 𝔟 is the ideal generated by a1,…,an,b1,…,bm.
(b)
𝔞 ⋅𝔟 is the ideal generated by aibj, for 1 ≤ i ≤ n and 1 ≤ j ≤ m.
Proof.Pretty straightforward; just convince yourself that this result is correct. □
In other words, for sums you append the two sets of generators together, and for products
you take products of the generators. Note that for principal ideals, this coincides with
“normal” multiplication, for example
in ℤ.
Remark 48.2.3 — Note that for an ideal 𝔞 and an element c, the set
is equal to (c) ⋅𝔞. So “scaling” and “multiplying by principal ideals” are the same thing.
This is important, since we’ll be using the two notions interchangably.
Finally, since we want to do factorization we better have some notion of divisibility. So we
define:
Definition 48.2.4.We say 𝔞 divides 𝔟 and write 𝔞∣𝔟 if 𝔞 ⊇𝔟.
Note the reversal of inclusions! So (3) divides (15), because (15) is contained in (3); every
multiple of 15 is a multiple of 3. And from the example in the previous section: In ℤ[],
(3,1 −) divides (3) and (1 −).
Finally, the prime ideals are defined as in ?? : 𝔭 is prime if xy ∈𝔭 implies x ∈𝔭 or y ∈𝔭.
This is compatible with the definition of divisibility:
Exercise 48.2.5.A nonzero proper ideal 𝔭 is prime if and only if whenever 𝔭 divides 𝔞𝔟, 𝔭divides one of 𝔞 or 𝔟.
As mentioned in ?? , this also lets us ignore multiplication by units: (−3) = (3).
48.3Dedekind domains
Prototypical example for this section:Any 𝒪Kis a Dedekind domain.
We now define a Dedekind domain as follows.
Definition 48.3.1.An integral domain 𝒜 is a Dedekind domain if it is Noetherian,
integrally closed, and every nonzero prime ideal of 𝒜 is in fact maximal. (The last
condition is the important one.)
Here there’s one new word I have to define for you, but we won’t make much use of it.
Definition 48.3.2.Let R be an integral domain and let K be its field of fractions.
We say R is integrally closed if the only elements a ∈ K which are roots of monic
polynomials in R are the elements of R (which are roots of the trivial x−r polynomial).
The interesting condition in the definition of a Dedekind domain is the last one: prime ideals
and maximal ideals are the same thing. The other conditions are just technicalities, but
“primes are maximal” has real substance.
Example 48.3.3 (ℤ is a Dedekind domain) The ring ℤ is a Dedekind domain. Note that
ℤ is Noetherian (for obvious reasons).
ℤ has field of fractions ℚ. If f(x) ∈ ℤ[x] is monic, then by the rational root theorem
any rational roots are integers (this is the same as the proof that ℤ∩ ℚ = ℤ).
Hence ℤ is integrally closed.
The nonzero prime ideals of ℤ are (p), which also happen to be maximal.
The case of interest is a ring 𝒪K in which we wish to do factorizing. We’re now going to
show that for any number field K, the ring 𝒪K is a Dedekind domain. First, the boring
part.
Proposition 48.3.4 (𝒪Kintegrally closed and Noetherian) For any number field K, the
ring 𝒪K is integrally closed and Noetherian.
Proof.Boring, but here it is anyways for completeness.
Since 𝒪Kℤ⊕n, we get that it’s Noetherian.
Now we show that 𝒪K is integrally closed. Suppose that η ∈ K is the root of some
polynomial with coefficients in 𝒪K. Thus
where αi∈𝒪K. We want to show that η ∈𝒪K as well.
Well, from the above, 𝒪K[η] is finitely generated…thus ℤ[η] ⊆ 𝒪K[η] is finitely
generated. So η ∈ℤ, and hence η ∈ K ∩ℤ = 𝒪K. □
Now let’s do the fun part. We’ll prove a stronger result, which will re-appear
repeatedly.
Theorem 48.3.5 (Important: prime ideals divide rational primes) Let 𝒪K be a ring of
integers and 𝔭 a nonzero prime ideal inside it. Then 𝔭 contains a rational prime p.
Moreover, 𝔭 is maximal.
Proof.Take any α≠0 in 𝔭. Its Galois conjugates are algebraic integers so their
product N(α)∕α is in 𝒪K (even though each individual conjugate need not be in K).
Consequently, N(α) ∈𝔭, and we conclude 𝔭 contains some integer.
Then take the smallest positive integer in 𝔭, say p. We must have that p is a rational
prime, since otherwise 𝔭 ∋ p = xy implies one of x,y ∈𝔭. This shows the first part.
We now do something pretty tricky to show 𝔭 is maximal. Look at 𝒪K∕𝔭; since 𝔭 is
prime it’s supposed to be an integral domain… but we claim that it’s actually finite! To
do this, we forget that we can multiply on 𝒪K. Recalling that 𝒪Kℤ⊕n as an abelian
group, we obtain a map
Hence ≤ pn is finite. Since finite integral domains are fields (?? ) we are
done. □
Since every nonzero prime 𝔭 is maximal, we now know that 𝒪K is a Dedekind domain. Note
that this tricky proof is essentially inspired by the solution to ?? .
48.4Unique factorization works
Okay, I’ll just say it now!
Unique factorization works perfectly in Dedekind domains!
Theorem 48.4.1 (Prime factorization works) Let 𝔞 be a nonzero proper ideal of a
Dedekind domain 𝒜. Then 𝔞 can be written as a finite product of nonzero prime ideals
𝔭i, say
and this factorization is unique up to the order of the 𝔭i.
Moreover, 𝔞 divides 𝔟 if and only if for every prime ideal 𝔭, the exponent of 𝔭 in 𝔞 is less
than the corresponding exponent in 𝔟.
I won’t write out the proof, but I’ll describe the basic method of attack. Section 3 of [?] does
a nice job of explaining it. When we proved the fundamental theorem of arithmetic, the basic
plot was:
(1)
Show that if p is a rational prime1
then p∣bc means p∣b or p∣c. (This is called Euclid’s Lemma.)
(2)
Use strong induction to show that every N > 1 can be written as the product of
primes (easy).
(3)
Show that if p1…pm = q1…qn for some primes (not necessarily unique), then p1 = qi
for some i, say q1.
(4)
Divide both sides by p1 and use induction.
What happens if we try to repeat the proof here? We get step 1 for free, because we’re using a
better definition of “prime”. We can also do step 3, since it follows from step 1. But step 2
doesn’t work, because for abstract Dedekind domains we don’t really have a notion of size.
And step 4 doesn’t work because we don’t yet have a notion of what the inverse of a prime
ideal is.
Well, it turns out that we can define the inverse 𝔞−1 of an ideal, and I’ll do so by the end of
this chapter. You then need to check that 𝔞 ⋅𝔞−1 = (1) = 𝒜. In fact, even this isn’t easy. You
have to check it’s true for prime ideals 𝔭, then prove prime factorization, and then prove that
this is true. Moreover, 𝔞−1 is not actually an ideal, so you need to work in the field of fractions
K instead of 𝒜.
So the main steps in the new situation are as follows:
(1)
First, show that every ideal 𝔞 divides 𝔭1…𝔭g for some finite collection of primes. (This
is an application of Zorn’s Lemma.)
(2)
Define 𝔭−1 and show that 𝔭𝔭−1 = (1).
(3)
Show that a factorization exists (again using Zorn’s Lemma).
(4)
Show that it’s unique, using the new inverse we’ve defined.
Finally, let me comment on how nice this is if 𝒜 is a PID (like ℤ). Thus every element a ∈𝒜
is in direct correspondence with an ideal (a). Now suppose (a) factors as a product of ideals
𝔭i = (pi), say,
This verbatim reads
where u is some unit (recall ?? ). Hence, Dedekind domains which are PID’s satisfy
unique factorization for elements, just like in ℤ. (In fact, the converse of this is
true.)
48.5The factoring algorithm
Let’s look at some examples from quadratic fields. Recall that if K = ℚ(), then
Also, recall that the norm of a + b is given by a2 + db2.
Example 48.5.1 (Factoring 6 in the integers of ℚ()) Let 𝒪K = ℤ[] arise from
K = ℚ(). We’ve already seen that
and you can’t get any further with these principal ideals. But let
Then it turns out (6) = 𝔭2𝔮1𝔮2. More specifically, (2) = 𝔭2, (3) = 𝔮1𝔮2, and (1+) =
𝔭𝔮1 and (1 −) = 𝔭𝔮2. (Proof in just a moment.)
I want to stress that all our ideals are computed relative to 𝒪K. So for example,
How do we know in this example that 𝔭 is prime/maximal? (Again, these are the same since
we’re in a Dedekind domain.) Answer: look at 𝒪K∕𝔭 and see if it’s a field. There is a trick to
this: we can express
So when we take that mod 𝔭, we get that
as rings.
Question 48.5.2.Conclude that 𝒪K∕𝔭𝔽2, and satisfy yourself that 𝔮1and 𝔮2are alsomaximal.
I should give an explicit example of an ideal multiplication: let’s compute
𝔮1𝔮2
=
=
=
= (3)
where we first did 9 − 6 = 3 (think Euclidean algorithm!), then noted that all the other
generators don’t contribute anything we don’t already have with the 3 (again these are
ideals computed in 𝒪K). You can do the computation for 𝔭2, 𝔭𝔮1, 𝔭𝔮2 in the same
way.
Finally, it’s worth pointing out that we should quickly verify that 𝔭≠(x) for some x; in other
words, that 𝔭 is not principal. Assume for contradiction that it is. Then x divides both
1 + and 2, in the sense that 1 + = α1x and 2 = α2x for some α1,α2∈𝒪K.
(Principal ideals are exactly the “multiples” of x, so (x) = x𝒪K.) Taking the norms, we find
that NK∕ℚ(x) divides both
Since 𝔭≠(1), x cannot be a unit, so its norm must be 2. But there are no elements of norm
2 = a2 + 5b2 in 𝒪K.
Example 48.5.3 (Factoring 3 in the integers of ℚ()) Let 𝒪K = ℤ[] arise
from K = ℚ(). We know 𝒪Kℤ[x]∕(x2 + 17). Now
This already shows that (3) cannot be a prime (i.e. maximal) ideal, since otherwise our
result should be a field. Anyways, we have a projection
Let 𝔮1 be the pre-image of (x − 1) in the image, that is,
Similarly,
We have 𝒪K∕𝔮1𝔽3, so 𝔮1 is maximal (prime). Similarly 𝔮2 is prime. Magically, you can
check explicitly that
Hence this is the factorization of (3) into prime ideals.
The fact that 𝔮1𝔮2 = (3) looks magical, but it’s really true:
𝔮1𝔮2
= (3,− 1)(3, + 1)
= (9,3 + 3,3− 3,18)
= (9,3 + 3,6)
= (3,3 + 3,6)
= (3).
In fact, it turns out this always works in general: given a rational prime p, there is an
algorithm to factor p in any 𝒪K of the form ℤ[𝜃].
Theorem 48.5.4 (Factoring algorithm / Dedekind-Kummer theorem) Let K be a number field.
Let 𝜃 ∈𝒪K with [𝒪K : ℤ[𝜃]] = j < ∞, and let p be a prime not dividing j. Then (p) = p𝒪K is
factored as follows:
Let f be the minimal polynomial of 𝜃 and factor f mod p as
Then 𝔭i = (fi(𝜃),p) is prime for each i and the factorization of (p) is
In particular, if K is monogenic with 𝒪K = ℤ[𝜃] then j = 1 and the theorem applies for all
primes p.
In almost all our applications in this book, K will be monogenic; i.e. j = 1. Here ψ denotes
the image in 𝔽p[x] of a polynomial ψ ∈ ℤ[x].
Question 48.5.5.There are many possible pre-images fiwe could have chosen (for example iffi= x2+ 1 (mod3), we could pick fi= x2+ 3x + 7.) Why does this not affect the value of 𝔭i?
Note that earlier, we could check the factorization worked for any particular case. The proof
that this works is much the same, but we need one extra tool, the ideal norm. After that we
leave the proposition as ?? .
This algorithm gives us a concrete way to compute prime factorizations of (p) in any
monogenic number field with 𝒪K = ℤ[𝜃]. To summarize the recipe:
1.
Find the minimal polynomial of 𝜃, say f ∈ ℤ[x].
2.
Factor f mod p into irreducible polynomials f1e1f2e2…fgeg.
3.
Compute 𝔭i = (fi(𝜃),p) for each i.
Then your (p) = 𝔭1e1…𝔭geg.
Exercise 48.5.6.Factor (29) in ℤ[i] using the above algorithm.
48.6Fractional ideals
Prototypical example for this section:Analog to ℚ for ℤ, allowing us to take inverses of ideals.Prime factorization works in the nicest way possible.
We now have a neat theory of factoring ideals of 𝒜, just like factoring the integers. Now
note that our factorization of ℤ naturally gives a way to factor elements of ℚ; just factor the
numerator and denominator separately.
Let’s make the analogy clearer. The analogue of a rational number is as follows.
Definition 48.6.1.Let 𝒜 be a Dedekind domain with field of fractions K. A fractionalidealJ of K is a set of the form
For emphasis, ideals of 𝒜 will be sometimes referred to as integral ideals.
You might be a little surprised by this definition: one would expect that a fractional ideal
should be of the form for some integral ideals 𝔞, 𝔟. But in fact, it suffices to just take x ∈𝒜
in the denominator. The analogy is that when we looked at 𝒪K, we found that we only needed
integer denominators: = (4 + ). Similarly here, it will turn out that we only need to
look at ⋅𝔞 rather than , and so we define it this way from the beginning. See ?? for a
different equivalent definition.
Example 48.6.2 (ℤ is a fractional ideal) The set
is a fractional ideal of ℤ.
Now, as we prescribed, the fractional ideals form a multiplicative group:
Theorem 48.6.3 (Fractional ideals form a group) Let 𝒜 be a Dedekind domain and K its
field of fractions. For any integral ideal 𝔞, the set
is a fractional ideal with 𝔞𝔞−1 = (1).
Definition 48.6.4.Thus nonzero fractional ideals of K form a group under
multiplication with identity (1) = 𝒜. This ideal group is denoted JK.
Example 48.6.5 ((3)−1 in ℤ) Please check that in ℤ we have
It follows that every fractional ideal J can be uniquely written as
where ni and mi are positive integers. In fact, 𝔞 is an integral ideal if and only if all its
exponents are nonnegative, just like the case with integers. So, a perhaps better way to
think about fractional ideals is as products of prime ideals, possibly with negative
exponents.
48.7The ideal norm
One last tool is the ideal norm, which gives us a notion of the “size” of an ideal.
Definition 48.7.1.The ideal norm (or absolute norm) of a nonzero ideal 𝔞 ⊆𝒪K is
defined as and denoted N(𝔞).
Example 48.7.2 (Ideal norm of (5) in the Gaussian integers) Let K = ℚ(i), 𝒪K =
ℤ[i]. Consider the ideal (5) in 𝒪K. We have that
so (5) has ideal norm 25, corresponding to the fact that 𝒪K∕(5) has 52 = 25 elements.
Example 48.7.3 (Ideal norm of (2 + i) in the Gaussian integers) You’ll notice that
since mod 2 + i we have both 5 ≡ 0 and i ≡−2. (Indeed, since (2 + i) is prime we had
better get a field!) Thus N = 5; similarly N = 5.
Thus the ideal norm measures how “roomy” the ideal is: that is, (5) is a lot more spaced out
in ℤ[i] than it is in ℤ. (This intuition will be important when we will actually view 𝒪K as a
lattice.)
Question 48.7.4.What are the ideals with ideal norm one?
Our example with (5) suggests several properties of the ideal norm which turn out to be
true:
Lemma 48.7.5 (Properties of the absolute norm) Let 𝔞 be a nonzero ideal of 𝒪K.
(a)
N(𝔞) is finite.
(b)
For any other nonzero ideal 𝔟, N(𝔞𝔟) = N(𝔞)N(𝔟).
(c)
If 𝔞 = (a) is principal, then N(𝔞) = NK∕ℚ(a).
I unfortunately won’t prove these properties, though we already did (a) in our proof that
𝒪K was a Dedekind domain.
The fact that N is completely multiplicative lets us also consider the norm of a fractional
ideal J by the natural extension
Thus N is a natural group homomorphism JK→ ℚ ∖{0}.
48.8A few harder problems to think about
Problem 48A.Show that there are three different factorizations of 77 in 𝒪K, where
K = ℚ().
Problem 48B.Let K = ℚ(); take for granted that 𝒪K = ℤ[]. Find the
factorization of (5) in 𝒪K.
Problem 48C (Fermat’s little theorem).Let 𝔭 be a prime ideal in some ring of integers
𝒪K. Show that for α ∈𝒪K,
Problem 48D†.Let 𝒜 be a Dedekind domain with field of fractions K, and pick J ⊆ K.
Show that J is a fractional ideal if and only if
(i)
J is closed under addition and multiplication by elements of 𝒜, and
(ii)
J is finitely generated as an abelian group.
More succinctly: J is a fractional ideal ⇔J is a finitely generated 𝒜-module.
Problem 48E. In the notation of ?? , let I = ∏i=1g𝔭iei. Assume for simplicity that K is
monogenic, hence 𝒪K = ℤ[𝜃].
(a)
Prove that each 𝔭i is prime.
(b)
Show that (p) divides I.
(c)
Use the norm to show that (p) = I.
49Minkowski bound and class groups
We now have a neat theory of unique factorization of ideals. In the case of a PID, this in fact
gives us a UFD. Sweet.
We’ll define, in a moment, something called the class group which measures how far 𝒪K is
from being a PID; the bigger the class group, the farther 𝒪K is from being a PID. In
particular, the 𝒪K is a PID if it has trivial class group.
Then we will provide some inequalities which let us put restrictions on the class group; for
instance, this will let us show in some cases that the class group must be trivial. Astonishingly,
the proof will use Minkowski’s theorem, a result from geometry.
49.1The class group
Prototypical example for this section:PID’s have trivial class group.
Let K be a number field, and let JK denote the multiplicative group of fractional ideals of
𝒪K. Let PK denote the multiplicative group of principal fractional ideals: those of the
form (x) = x𝒪K for some x ∈ K.
Question 49.1.1.Check that PKis also a multiplicative group. (This is really easy: namex𝒪K⋅ y𝒪Kand (x𝒪K)−1.)
As JK is abelian, we can now define the class group to be the quotient
The elements of ClK are called classes.
Equivalently,
The class group ClKis the set of nonzero fractional ideals modulo scaling by aconstant in K.
In particular, ClK is trivial if all ideals are principal, since the nonzero principal ideals are
the same up to scaling.
The size of the class group is called the class number. It’s a beautiful theorem that the
class number is always finite, and the bulk of this chapter will build up to this result. It
requires several ingredients.
49.2The discriminant of a number field
Prototypical example for this section:Quadratic fields.
Let’s say I have K = ℚ(). As we’ve seen before, this means 𝒪K = ℤ[], meaning
The key insight now is that you might think of this as a lattice: geometrically, we want to
think about this the same way we think about ℤ2.
Perversely, we might try to embed this into ℚ2 by sending a + b to (a,b). But this is a
little stupid, since we’re rudely making K, which somehow lives inside ℝ and is
“one-dimensional” in that sense, into a two-dimensional space. It also depends on a
choice of basis, which we don’t like. A better way is to think about the fact that
there are two embeddings σ1 : K → ℂ and σ2 : K → ℂ, namely the identity, and
conjugation:
σ1(a + b)
= a + b
σ2(a + b)
= a − b.
Fortunately for us, these embeddings both have real image. This leads us to consider the set
of points
This lets us visualize what 𝒪K looks like in ℝ2. The points of K are dense in ℝ2, but the
points of 𝒪K cut out a lattice.
To see how big the lattice is, we look at how {1,}, the generators of 𝒪K, behave. The
point corresponding to a + b in the lattice is
The mesh of the lattice1
is defined as the hypervolume of the “fundamental parallelepiped” I’ve colored blue above. For
this particular case, it ought to be equal to the area of that parallelogram, which
is
The definition of the discriminant is precisely this, except with an extra square factor (since
permutation of rows could lead to changes in sign in the matrix above). ?? shows that the
squaring makes ΔK an integer.
To make the next definition, we invoke:
Theorem 49.2.1 (The n embeddings of a number field) Let K be a number field of degree
n. Then there are exactly n field homomorphisms Kℂ, say σ1,…,σn, which fix ℚ.
Proof.Deferred to ?? , once we have the tools of Galois theory. □
In fact, in ?? we see that for α ∈ K, we have that σi(α) runs over the conjugates of α as
i = 1,…,n. It follows that
This allows us to define:
Definition 49.2.2.Suppose α1,…,αn is a ℤ-basis of 𝒪K. The discriminant of the
number field K is defined by
This does not depend on the choice of the {αi}; we will not prove this here.
Example 49.2.3 (Discriminant of K = ℚ()) We have 𝒪K = ℤ[] and as discussed
above the discriminant is
Example 49.2.4 (Discriminant of ℚ(i)) Let K = ℚ(i). We have 𝒪K = ℤ[i] = ℤ ⊕ iℤ.
The embeddings are the identity and complex conjugation which take 1 to (1,1) and i
to (i,−i). So
This example illustrates that the discriminant need not be positive for number fields
which wander into the complex plane (the lattice picture is a less perfect analogy). But
again, as we’ll prove in the problems the discriminant is always an integer.
Example 49.2.5 (Discriminant of ℚ()) Let K = ℚ(). This time, 𝒪K = ℤ ⊕ℤ, and so the discriminant is going to look a little bit different. The embeddings
are still a + ba + b,a − b.
Applying this to the ℤ-basis , we get
Exercise 49.2.6.Extend all this to show that if K = ℚ() for d≠1 squarefree, we have
Actually, let me point out something curious: recall that the polynomial discriminant of
Ax2 + Bx + C is B2− 4AC. Then:
In the d ≡ 1 (mod 4) case, ΔK is the discriminant of x2− x −, which is the
minimal polynomial of (1 + ). Of course, 𝒪K = ℤ[(1 + )].
In the d ≡ 2,3 (mod 4) case, ΔK is the discriminant of x2−d which is the minimal
polynomial of . Once again, 𝒪K = ℤ[].
This is not a coincidence! ?? asserts that this is true in general; hence the name
“discriminant”.
49.3The signature of a number field
Prototypical example for this section:ℚ() has signature (2,49).
In the example of K = ℚ(i), we more or less embedded K into the space ℂ. However, K is a
degree two extension, so what we’d really like to do is embed it into ℝ2. To do so, we’re going
to take advantage of complex conjugation.
Let K be a number field and σ1,…,σn be its embeddings. We distinguish between the realembeddings (which map all of K into ℝ) and the complex embeddings (which map some
part of K outside ℝ). Notice that if σ is a complex embedding, then so is the conjugate σ≠σ;
hence complex embeddings come in pairs.
Definition 49.3.1.Let K be a number field of degree n, and set
r1
= number of real embeddings
r2
= number of pairs of complex embeddings.
The signature of K is the pair (r1,r2). Observe that r1 + 2r2 = n.
Example 49.3.2 (Basic examples of signatures)
(a)
ℚ has signature (1,0).
(b)
ℚ() has signature (2,0).
(c)
ℚ(i) has signature (0,1).
(d)
Let K = ℚ(), and let ω be a cube root of unity. The elements of K are
Then the signature is (1,1), because the three embeddings are
The first of these is real and the latter two are conjugate pairs.
Example 49.3.3 (Even more signatures) In the same vein ℚ() and ℚ() have
signatures (1,49) and (2,49).
Question 49.3.4.Verify the signatures of the above two number fields.
From now on, we will number the embeddings of K in such a way that
are the real embeddings, while
are the r2 pairs of complex embeddings. We define the canonical embedding of K
as
All we’ve done is omit, for the complex case, the second of the embeddings in each conjugate
pair. This is no big deal, since they are just conjugates; the above tuple is all the information
we need.
For reasons that will become obvious in a moment, I’ll let τ denote the isomorphism
by breaking each complex number into its real and imaginary part, as
α
σ1(α),…,σr1(α),
Reσr1+1(α), Imσr1+1(α),
Reσr1+2(α), Imσr1+2(α),
…,
Reσr1+r2(α), Imσr1+r2(α).
Example 49.3.5 (Example of canonical embedding) As before let K = ℚ() and set
where ω = − +i, noting that we’ve already arranged indices so σ1 = id is real while
σ2 and σ3 are a conjugate pair. So the embeddings Kℝ × ℂℝ3 are given by
For concreteness, taking α = 9 + gives
9 +
= ∈ ℝ × ℂ
∈ ℝ3.
Now, the whole point of this is that we want to consider the resulting lattice when we take
𝒪K. In fact, we have:
Lemma 49.3.6 Consider the composition of the embeddings Kℝr1× ℂr2ℝn. Then
as before, 𝒪K becomes a lattice L in ℝn, with mesh equal to
Proof.Fun linear algebra problem (you just need to manipulate determinants). Left as
?? . □
From this we can deduce:
Lemma 49.3.7 Consider the composition of the embeddings Kℝr1× ℂr2ℝn. Let 𝔞
be an ideal in 𝒪K. Then the image of 𝔞 is a lattice L𝔞 in ℝn with mesh equal to
Sketch of Proof. Let
Then in the lattice L𝔞, we somehow only take th of the points which appear in the
lattice L, which is why the area increases by a factor of N(𝔞). To make this all precise
I would need to do a lot more with lattices and geometry than I have space for in this
chapter, so I will omit the details. But I hope you can see why this is intuitively true. □
49.4Minkowski’s theorem
Now I can tell you why I insisted we move from ℝr1× ℂr2 to ℝn. In geometry, there’s a really
cool theorem of Minkowski’s that goes as follows.
Theorem 49.4.1 (Minkowski) Let S ⊆ ℝn be a convex set containing 0 which is centrally
symmetric (meaning that x ∈ S⇔− x ∈ S). Let L be a lattice with mesh d. If
either
(a)
The volume of S exceeds 2nd, or
(b)
The volume of S equals 2nd and S is compact,
then S contains a nonzero lattice point of L.
Question 49.4.2.Show that the condition 0 ∈ S is actually extraneous in the sense that anynonempty, convex, centrally symmetric set contains the origin.
Sketch of Proof. Part (a) is surprisingly simple and has a very olympiad-esque solution:
it’s basically Pigeonhole on areas. We’ll prove part (a) in the special case n = 2, L = ℤ2
for simplicity as the proof can easily be generalized to any lattice and any n. Thus we
want to show that any such convex set S with area more than 4 contains a lattice point.
Dissect the plane into 2 × 2 squares
and overlay all these squares on top of each other. By the Pigeonhole Principle, we
find there exist two points p≠q ∈ S which map to the same point. Since S is symmetric,
−q ∈ S. Then (p − q) ∈ S (convexity) and is a nonzero lattice point.
I’ll briefly sketch part (b): the idea is to consider (1+𝜀)S for 𝜀 > 0 (this is “S magnified
by a small factor 1+𝜀”). This satisfies condition (a). So for each 𝜀 > 0 the set of nonzero
lattice points in (1+𝜀)S, say S𝜀, is a finite nonempty set of (discrete) points (the “finite”
part follows from the fact that (1+𝜀)S is bounded). So there has to be some point that’s
in S𝜀 for every 𝜀 > 0 (why?), which implies it’s in S. □
49.5The trap box
The last ingredient we need is a set to apply Minkowski’s theorem to. I propose:
Definition 49.5.1.Let M be a positive real. In ℝr1× ℂr2, define the box S to be the
set of points (x1,…,xr1,z1,…,zr2) such that
Note that this depends on the value of M.
Think of this box as a mousetrap: anything that falls in it is going to have a
small norm, and our goal is to use Minkowski to lure some nonzero element into
it.
That is, suppose α ∈𝔞 falls into the box I’ve defined above, which means
where we are remembering that the last few σ’s come in conjugate pairs. This looks like
the trace, but the absolute values are in the way. So instead, we apply AM-GM to
obtain:
Lemma 49.5.2 (Effect of the mousetrap) Let α ∈ 𝒪K, and suppose ι(α) is in S (where
ι : Kℝr1× ℂr2 as usual). Then
The last step we need to do is compute the volume of the box. This is again some geometry
I won’t do, but take my word for it:
Lemma 49.5.3 (Size of the mousetrap) Let τ : ℝr1× ℂr2ℝn as before. Then the image
of S under τ is a convex, compact, centrally symmetric set with volume
Question 49.5.4.(Sanity check) Verify that the above is correct for the signatures (r1,r2) =(2,0) and (r1,r2) = (0,1), which are the possible signatures when n = 2.
49.6The Minkowski bound
We can now put everything we have together to obtain the great Minkowski bound.
Theorem 49.6.1 (Minkowski bound) Let 𝔞 ⊆𝒪K be any nonzero ideal. Then there exists
0≠α ∈𝔞 such that
Proof.This is a matter of putting all our ingredients together. Let’s see what things we’ve
defined already:
Pick a value of M such that the mesh of L𝔞 equals 2−n of the volume of the box. Then
Minkowski’s theorem gives that some 0≠α ∈𝔞 lands inside the box — the mousetrap is
configured to force NK∕ℚ(α) ≤Mn. The correct choice of M is
which gives the bound after some arithmetic. □
49.7The class group is finite
Definition 49.7.1.Let MK = r2 for brevity. Note that it is a constant
depending on K.
So that’s cool and all, but what we really wanted was to show that the class group is finite.
How can the Minkowski bound help? Well, you might notice that we can rewrite it to
say
where MK is some constant depending on K, and α ∈𝔞.
Question 49.7.2.Show that (α) ⋅𝔞−1is an integral ideal. (Unwind definitions.)
But in the class group we mod out by principal ideals like (α). If we shut our eyes for a
moment and mod out, the above statement becomes “N(𝔞−1) ≤ MK”. The precise statement
of this is
Corollary 49.7.3 Let K be a number field, and pick a fractional ideal J. Then we can
find α such that 𝔟 = (α) ⋅ J is integral and N(𝔟) ≤ MK.
Proof.For fractional ideals I and J write I ∼ J to mean that I = (α)J for some α;
then ClK is just modding out by ∼. Let J be a fractional ideal. Then J−1 is some other
fractional ideal. By definition, for some α ∈𝒪K we have that αJ−1 is an integral ideal
𝔞. The Minkowski bound tells us that for some x ∈ 𝔞, we have N(x𝔞−1) ≤ MK. But
x𝔞−1∼𝔞−1 = (αJ−1)−1∼ J. □
Corollary 49.7.4 (Finiteness of class group) Class groups are always finite.
Proof.For every class in ClK, we can identify an integral ideal 𝔞 with norm less than
MK. We just have to show there are finitely many such integral ideals; this will mean
there are finitely many classes.
Suppose we want to build such an ideal 𝔞 = 𝔭1e1…𝔭mem. Recall that a prime ideal 𝔭i must
have some rational prime p inside it, meaning 𝔭i divides (p) and p divides N(𝔭i). So
let’s group all the 𝔭i we want to build 𝔞 with based on which (p) they came from.
To be more dramatic: imagine you have a cherry tree; each branch corresponds to a prime (p)
and contains as cherries (prime ideals) the factors of (p) (finitely many). Your bucket (the
ideal 𝔞 you’re building) can only hold a total weight (norm) of MK. So you can’t even touch
the branches higher than MK. You can repeat cherries (oops), but the weight of a cherry on
branch (p) is definitely ≥ p; all this means that the number of ways to build 𝔞 is
finite. □
49.8Computation of class numbers
Definition 49.8.1.The order of ClK is called the class number of K.
Remark 49.8.2 — If ClK = 1, then 𝒪K is a PID, hence a UFD.
By computing the actual value of MK, we can quite literally build the entire “cherry tree”
mentioned in the previous proof. Let’s give an example how!
Proposition 49.8.3 The field ℚ() has class number 1.
Proof.Since K = ℚ() has signature (0,1) and discriminant ΔK = −67 (since
−67 ≡ 1 (mod 4)) we can compute
That means we can cut off the cherry tree after (2), (3), (5), since any cherries on these
branches will necessarily have norm ≥ MK. We now want to factor each of these in
𝒪K = ℤ[𝜃], where 𝜃 = has minimal polynomial x2−x + 17. But something miraculous
happens:
When we try to reduce x2− x + 17 (mod 2), we get an irreducible polynomial
x2− x + 1. By the factoring algorithm (?? ) this means (2) is prime.
Similarly, reducing mod 3 gives x2− x + 2, which is irreducible. This means (3) is
prime.
Finally, for the same reason, (5) is prime.
It’s our lucky day; all of the ideals (2), (3), (5) are prime (already principal). To put it another
way, each of the three branches has only one (large) cherry on it. That means any time we put
together an integral ideal with norm ≤ MK, it is actually principal. In fact, these guys have
norm 4, 9, 25 respectively…so we can’t even touch (3) and (5), and the only ideals we can get
are (1) and (2) (with norms 1 and 4).
Now we claim that’s all. Pick a fractional ideal J. By ?? , we can find an integral ideal
𝔟 ∼ J with N(𝔟) ≤ MK. But by the above, either 𝔟 = (1) or 𝔟 = (2), both of which are
principal, and hence trivial in ClK. So J is trivial in ClK too, as needed. □
Let’s do a couple more.
Theorem 49.8.4 (Gaussian integers ℤ[i] form a UFD) The field ℚ(i) has class number 1.
Proof.This is 𝒪K where K = ℚ(i), so we just want ClK to be trivial. We have
MK = < 2. So every class has an integral ideal of norm 𝔟 satisfying
Well, that’s silly: we don’t have any branches to pick from at all. In other words, we
can only have 𝔟 = (1). □
Here’s another example of something that still turns out to be unique factorization, but this
time our cherry tree will actually have cherries that can be picked.
Proposition 49.8.5 (ℤ[] is a UFD) The field ℚ() has class number 1.
Proof.First we compute the Minkowski bound.
Question 49.8.6.Check that MK≈ 2.646.
So this time, the only branch is (2). Let’s factor (2) as usual: the polynomial x2 + 7 reduces
as (x − 1)(x + 1) (mod 2), and hence
Oops! We now have two cherries, and they both seem reasonable. But actually, I claim
that
Question 49.8.7.Prove this.
So both the cherries are principal ideals, and as before we conclude that ClK is trivial. But
note that this time, the prime ideal (2) actually splits; we got lucky that the two cherries were
principal but this won’t always work. □
How about some nontrivial class groups? First, we use a lemma that will help us with
narrowing down the work in our cherry tree.
Lemma 49.8.8 (Ideals divide their norms) Let 𝔟 be an integral ideal with N(𝔟) = n. Then
𝔟 divides the ideal (n).
Proof.By definition, n = . Treating 𝒪K∕𝔟 as an (additive) abelian group and
using Lagrange’s theorem, we find
Thus (n) ⊆𝔟, done. □
Now we can give such an example.
Proposition 49.8.9 (Class group of ℚ()) The number field K = ℚ() has class
group ℤ∕4ℤ.
You are not obliged to read the entire proof in detail, as it is somewhat gory. The idea is
just that there are some cherries which are not trivial in the class group.
Proof.Since ΔK = −68, we compute the Minkowski bound
Now, it suffices to factor with (2), (3), (5). The minimal polynomial of is x2 + 17, so
as usual
(2)
= (2, + 1)2
(3)
= (3,− 1)(3, + 1)
(5)
= (5)
corresponding to the factorizations of x2 + 17 modulo each of 2, 3, 5. Set 𝔭 = (2, + 1)
and 𝔮1 = (3,− 1), 𝔮2 = (3, + 1). We can compute
In particular, they are not principal. The ideal (5) is out the window; it has norm 25. Hence,
the three cherries are 𝔭, 𝔮1, 𝔮2.
The possible ways to arrange these cherries into ideals with norm ≤ 5 are
However, you can compute
so 𝔭2 and (1) are in the same class group; that is, they are trivial. In particular, the class
group has order at most 4.
From now on, let [𝔞] denote the class (member of the class group) that 𝔞 is in. Since 𝔭 isn’t
principal (so [𝔭]≠[(1)]), it follows that 𝔭 has order two. So Lagrange’s theorem says that ClK
has order either 2 or 4.
Now we claim [𝔮1]2≠[(1)], which implies that 𝔮1 has order greater than 2. If not, 𝔮12 is
principal. We know N(𝔮1) = 3, so this can only occur if 𝔮12 = (3); this would force 𝔮1 = 𝔮2.
This is impossible since 𝔮1 + 𝔮2 = (1).
Thus, 𝔮1 has even order greater than 2. So it has to have order 4. From this we
deduce
Remark 49.8.10 — When we did this at Harvard during Math 129, there was a five-minute
interruption in which students (jokingly) complained about the difficulty of evaluating .
Excerpt:
“Will we be allowed to bring a small calculator on the exam?” – Student 1 “What does the size have to do with anything? You could have an Apple
Watch” – Professor “Just use the fact that π ≥ 3” – me “Even [other professor] doesn’t know that, how are we supposed to?” –
Student 2 “You have to do this yourself!” – Professor “This is an outrage.” – Student 1
49.9A few harder problems to think about
Problem 49A.Show that K = ℚ() has trivial class group, and hence 𝒪K = ℤ
is a UFD.2
Problem 49B.Determine the class group of ℚ().
Problem 49C (China TST 1998).Let n be a positive integer. A polygon in the plane (not
necessarily convex) has area greater than n. Prove that one can translate it so that it contains
at least n + 1 lattice points.
Problem 49D (?? ). Consider the composition of the embeddings Kℝr1×ℂr2ℝn. Show that the image of 𝒪K⊆ K has mesh equal to
Problem 49E.Let p ≡ 1 (mod 4) be a prime. Show that there are unique integers a > b > 0
such that a2 + b2 = p.
Problem 49F (Korea national olympiad 2014).Let p be an odd prime and k a positive
integer such that p∣k2 + 5. Prove that there exist positive integers m, n such that
p2 = m2 + 5n2.
50More properties of the discriminant
I’ll remind you that the discriminant of a number field K is given by
where α1, …, αn is a ℤ-basis for K, and the σi are the n embeddings of K into
ℂ.
Several examples, properties, and equivalent definitions follow.
50.1A few harder problems to think about
Problem 50A⋆ (Discriminant of cyclotomic field). Let p be an odd rational prime and
ζp a primitive pth root of unity. Let K = ℚ(ζp). Show that
Problem 50B⋆ (Trace representation of ΔK).Let α1, …, αn be a
basis for 𝒪K. Prove that
In particular, ΔK is an integer.
Problem 50C⋆ (Root representation of ΔK).The discriminant of a quadratic
polynomial Ax2 + Bx + C is defined as B2− 4AC. More generally, the polynomial
discriminant of a polynomial f ∈ ℤ[x] of degree n is
where z1,…,zn are the roots of f, and c is the leading coefficient of f.
Suppose K is monogenic with 𝒪K = ℤ[𝜃]. Let f denote the minimal polynomial of 𝜃
(hence monic). Show that
Problem 50D.Show that if K≠ℚ is a number field then > 1.
Problem 50E (Brill’s theorem).For a number field K with signature (r1,r2), show
that ΔK> 0 if and only if r2 is even.
Problem 50F (Stickelberger theorem).Let K be a number field.
Prove that
51Bonus: Let’s solve Pell’s equation!
This is an optional aside, and can be safely ignored. (On the other hand, it’s pretty
short.)
51.1Units
Prototypical example for this section:±1, roots of unity, 3 − 2and its powers.
Recall according to ?? that α ∈𝒪K is invertible if and only if
We let 𝒪K× denote the set of units of 𝒪K.
Question 51.1.1.Show that 𝒪K×is a group under multiplication. Hence we name it the unitgroupof 𝒪K.
What are some examples of units?
Example 51.1.2 (Examples of units in a number field)
1.
±1 are certainly units, present in any number field.
2.
If 𝒪K contains a root of unity ω (i.e. ωn = 1), then ω is a unit. (In fact, ±1 are
special cases of this.)
3.
Of course, not all units of 𝒪K are roots of unity. For example, if 𝒪K = ℤ[]
(from K = ℚ()) then the number 2 + is a unit, as its norm is
Alternatively, just note that the inverse 2 −∈𝒪K as well:
Either way, 2 − is a unit.
4.
Given any unit u ∈𝒪K×, all its powers are also units. So for example, (3 − 2)n
is always a unit of ℤ[], for any n. If u is not a root of unity, then this generates
infinitely many new units in 𝒪K×.
Question 51.1.3.Verify the claims above that
(a)
Roots of unity are units, and
(b)
Powers of units are units.
One can either proceed from the definition or use the characterizationNK∕ℚ(α) = ±1. If one definitionseems more natural to you, use the other.
51.2Dirichlet’s unit theorem
Prototypical example for this section:The units of ℤ[] are ±(2 + )n.
Definition 51.2.1.Let μ(𝒪K) denote the set of roots of unity contained in a number
field K (equivalently, in 𝒪K).
Example 51.2.2 (Examples of μ(𝒪K))
(a)
If K = ℚ(i), then 𝒪K = ℤ[i]. So
(b)
If K = ℚ(), then 𝒪K = ℤ[]. So
(c)
If K = ℚ(), then 𝒪K = ℤ[(1 + )]. So
where the ±’s in the second term need not depend on each other; in other words
μ(𝒪K) = .
Exercise 51.2.3.Show that we always have that μ(𝒪K) comprises the roots to xn−1 for someinteger n. (First, show it is a finite group under multiplication.)
We now quote, without proof, the so-called Dirichlet’s unit theorem, which gives
us a much more complete picture of what the units in 𝒪K are. Legend says that
Dirichlet found the proof of this theorem during an Easter concert in the Sistine
Chapel.
Theorem 51.2.4 (Dirichlet’s unit theorem) Let K be a number field with signature (r1,r2)
and set
Then there exist units u1, …, us such that every unit α ∈𝒪K× can be written uniquely
in the form
for ω ∈ μ(𝒪K) is a root of unity, and n1,…,ns∈ ℤ.
More succinctly:
We have 𝒪K×ℤr1+r2−1× μ(𝒪K).
A choice of u1, …, us is called a choice of fundamental units.
Here are some example applications.
Example 51.2.5 (Some unit groups)
(a)
Let K = ℚ(i) with signature (0,1). Then we obtain s = 0, so Dirichlet’s Unit
theorem says that there are no units other than the roots of unity. Thus
This is not surprising, since a + bi ∈ ℤ[i] is a unit if and only if a2 + b2 = 1.
(b)
Let K = ℚ(), which has signature (2,0). Then s = 1, so we expect exactly one
fundamental unit. A fundamental unit is 2 + (or 2 −, its inverse) with norm
1, and so we find
(c)
Let K = ℚ() with signature (1,1). Then s = 1, so we expect exactly one
fundamental unit. The choice 1 + + . So
I haven’t actually shown you that these are fundamental units, and indeed computing
fundamental units is in general hard.
51.3Finding fundamental units
Here is a table with some fundamental units.
In general, determining fundamental units is computationally hard.
However, once I tell you what the fundamental unit is, it’s not too bad (at least in the case
s = 1) to verify it. For example, suppose we want to show that 10 + 3 is a fundamental
unit of K = ℚ(1), which has ring of integers ℤ[]. If not, then for some n > 1, we would
have to have
For this to happen, at the very least we would need < 3. We would also have
x2− 11y2 = ±1. So one can just verify (using y = 1,2) that this fails.
The point is that: Since (10,3) is the smallest (in the sense of ) integer solution to
x2− 11y2 = ±1, it must be the fundamental unit. This holds more generally, although in the
case that d ≡ 1 (mod 4) a modification must be made as x, y might be half-integers (like
(1 + )).
Theorem 51.3.1 (Fundamental units of pell equations) Assume d is a squarefree integer.
(a)
If d ≡ 2,3 (mod 4), and (x,y) is a minimal integer solution to x2− dy2 = ±1, then
x + y is a fundamental unit.
(b)
If d ≡ 1 (mod 4), and (x,y) is a minimal half-integer solution to x2− dy2 = ±1,
then x + y is a fundamental unit. (Equivalently, the minimal integer solution to
a2− db2 = ±4 gives (a + b).)
(Any reasonable definition of “minimal” will work, such as sorting by .)
51.4Pell’s equation
This class of results completely eradicates Pell’s Equation. After all, solving
amounts to finding elements of ℤ[] with norm ±1. It’s a bit weirder in the d ≡ 1 (mod 4)
case, since in that case K = ℚ() gives 𝒪K = ℤ[(1 + )], and so the fundamental unit
may not actually be a solution. (For example, when d = 5, we get the solution (,).)
Nonetheless, all integer solutions are eventually generated.
To make this all concrete, here’s a simple example.
Example 51.4.1 (x2− 5y2 = ±1) Set K = ℚ(), so 𝒪K = ℤ[(1+)]. By Dirichlet’s
unit theorem, 𝒪K× is generated by a single element u. The choice
serves as a fundamental unit, as there are no smaller integer solutions to a2− 5b2 = ±4.
The first several powers of u are
One can see that the first integer solution is (2,1), which gives −1. The first solution
with +1 is (9,4). Continuing the pattern, we find that every third power of u gives an
integer solution (see also ?? ), with the odd ones giving a solution to x2− 5y2 = −1
and the even ones a solution to x2− 5y2 = +1. All solutions are generated this way, up
to ± signs (by considering ±u±n).
51.5A few harder problems to think about
Problem 51A (Fictitious account of the battle of Hastings).Determine the number of
soldiers in the following battle:
The men of Harold stood well together, as their wont was, and formed
thirteen squares, with a like number of men in every square thereof, and woe
to the hardy Norman who ventured to enter their redoubts; for a single blow
of Saxon war-hatched would break his lance and cut through his coat of mail
. . . when Harold threw himself into the fray the Saxons were one might
square of men, shouting the battle-cries, “Ut!”, “Olicrosse!”, “Godemite!”
Problem 51B. Let d > 0 be a squarefree integer, and let u denote the fundamental
unit of ℚ(). Show that either u ∈ ℤ[], or un∈ ℤ[] ⇔ 3∣n.
Problem 51C.Show that there are no integer solutions to
despite the fact that −1 is a quadratic residue mod 34.
Part XIV Algebraic NT II: Galois and Ramification Theory
52Things Galois
52.1Motivation
Prototypical example for this section:ℚ() and ℚ().
The key idea in Galois theory is that of embeddings, which give us another way to get at the
idea of the “conjugate” we described earlier.
Let K be a number field. An embeddingσ: Kℂ, is an injective field homomorphism: it
needs to preserve addition and multiplication, and in particular it should fix 1.
Question 52.1.1.Show that in this context, σ(q) = q for any rational number q.
Example 52.1.2 (Examples of embeddings)
(a)
If K = ℚ(i), the two embeddings of K into ℂ are zz (the identity) and zz
(complex conjugation).
(b)
If K = ℚ(), the two embeddings of K into ℂ are a+ba+b (the identity)
and a + ba − b (conjugation).
(c)
If K = ℚ(), there are three embeddings:
The identity embedding, which sends 11 and .
An embedding which sends 11 and ω, where ω is a cube root of
unity. Note that this is enough to determine the rest of the embedding.
An embedding which sends 11 and ω2.
I want to make several observations about these embeddings, which will form the core ideas
of Galois theory. Pay attention here!
First, you’ll notice some duality between roots: in the first example, i gets sent to
±i, gets sent to ±, and gets sent to the other roots of x3− 2. This is no
coincidence, and one can show this occurs in general. Specifically, suppose α has
minimal polynomial
where the ci are rational. Then applying any embedding σ to both sides gives
where in the last step we have used the fact that ci∈ ℚ, so they are fixed by σ. So, rootsof minimal polynomials go to other roots of that polynomial.
Next, I want to draw out a contrast between the second and third examples. Specifically,
in example (b) where we consider embeddings K = ℚ() to ℂ. The image
of these embeddings lands entirely in K: that is, we could just as well have
looked at K → K rather than looking at K → ℂ. However, this is not true
in (c): indeed ℚ() ⊂ ℝ, but the non-identity embeddings have complex
outputs!
The key difference is to again think about conjugates. Key observation:
The field K = ℚ() is “deficient” because the minimal polynomial x3− 2 hastwo other roots ωand ω2not contained in K.
On the other hand K = ℚ() is just fine because both roots of x2− 2 are contained
inside K. Finally, one can actually fix the deficiency in K = ℚ() by completing it to a
field ℚ(,ω). Fields like ℚ(i) or ℚ() which are “self-contained” are called Galoisextensions, as we’ll explain shortly.
Finally, you’ll notice that in the examples above, the number of embeddingsfrom K to ℂ happens to be the degree of K. This is an important theorem,
?? .
In this chapter we’ll develop these ideas in full generality, for any field other than
ℚ.
52.2Field extensions, algebraic closures, and splitting fields
Prototypical example for this section:ℚ()∕ℚ is an extension, ℂ is an algebraic closure ofany number field.
First, we define a notion of one field sitting inside another, in order to generalize the notion
of a number field.
Definition 52.2.1.Let K and F be fields. If F ⊆ K, we write K∕F and say K is a
field extension of F.
Thus K is automatically an F-vector space (just like ℚ() is automatically a ℚ-vector
space). The degree is the dimension of this space, denoted [K : F]. If [K : F] is finite,
we say K∕F is a finite (field) extension.
That’s really all. There’s nothing tricky at all.
Question 52.2.2.What do you call a finite extension of ℚ?
Degrees of finite extensions are multiplicative.
Theorem 52.2.3 (Field extensions have multiplicative degree) Let F ⊆ K ⊆ L be fields
with L∕K, K∕F finite. Then
Proof.Basis bash: you can find a basis of L over K, and then expand that into a basis
L over F. (Diligent readers can fill in details.) □
Next, given a field (like ℚ()) we want something to embed it into (in our case ℂ). So we
just want a field that contains all the roots of all the polynomials:
Theorem 52.2.4 (Algebraic closures) Let F be a field. Then there exists a field extension
F containing F, called an algebraic closure, such that all polynomials in F[x] factor
completely.
Example 52.2.5 (ℂ) ℂ is an algebraic closure of ℚ, ℝ and even itself.
Abuse of Notation 52.2.6.Some authors also require the algebraic closure to be
minimal by inclusion: for example, given ℚ they would want only ℚ (the algebraic
numbers). It’s a theorem that such a minimal algebraic closure is unique, and so these
authors will refer to the algebraic closure of K.
I like ℂ, so I’ll use the looser definition.
52.3Embeddings into algebraic closures for number fields
Now that I’ve defined all these ingredients, I can prove:
Theorem 52.3.1 (The n embeddings of a number field) Let K be a number field of degree
n. Then there are exactly n field homomorphisms Kℂ, say σ1,…,σn which fix ℚ.
Remark 52.3.2 — Note that a nontrivial homomorphism of fields is necessarily injective
(the kernel is an ideal). This justifies the use of “”, and we call each σi an embedding
of K into ℂ.
Proof.This is actually kind of fun! Recall that any irreducible polynomial over ℚ has
distinct roots (?? ). We’ll adjoin elements α1,α2,…,αm one at a time to ℚ, until we
eventually get all of K, that is,
Diagrammatically, this is
First, we claim there are exactly
ways to pick τ1. Observe that τ1 is determined by where it sends α1 (since it has to fix ℚ).
Letting p1 be the minimal polynomial of α1, we see that there are deg p1 choices for τ1, one for
each (distinct) root of p1. That proves the claim.
Similarly, given a choice of τ1, there are
ways to pick τ2. (It’s a little different: τ1 need not be the identity. But it’s still true that τ2
is determined by where it sends α2, and as before there are [ℚ(α1,α2) : ℚ(α1)] possible
ways.)
Multiplying these all together gives the desired [K : ℚ]. □
Remark 52.3.3 — The primitive element theorem actually implies that m = 1 is
sufficient; we don’t need to build a whole tower. This simplifies the proof somewhat.
It’s common to see expressions like “let K be a number field of degree n, and σ1,…,σn its n
embeddings” without further explanation. The relation between these embeddings and the
Galois conjugates is given as follows.
Theorem 52.3.4 (Embeddings are evenly distributed over conjugates) Let K be a number
field of degree n with n embeddings σ1, …, σn, and let α ∈ K have m Galois conjugates
over ℚ.
Then σj(α) is “evenly distributed” over each of these m conjugates: for any Galois
conjugate β, exactly of the embeddings send α to β.
Proof.In the previous proof, adjoin α1 = α first. □
So, now we can define the trace and norm over ℚ in a nice way: given a number field K, we
set
where σi are the n embeddings of K into ℂ.
52.4Everyone hates characteristic 2: separable vs irreducible
Prototypical example for this section:ℚ has characteristic zero, hence irreducible polynomialsare separable.
Now, we want a version of the above theorem for any field F. If you read the proof, you’ll
see that the only thing that ever uses anything about the field ℚ is ?? , where we use the fact
that
Irreducible polynomials over F have no double roots.
Let’s call a polynomial with no double roots separable; thus we want irreducible
polynomials to be separable. We did this for ℚ in the last chapter by taking derivatives.
Should work for any field, right?
Nope. Suppose we took the derivative of some polynomial like 2x3 + 24x + 9, namely
6x2 + 24. In ℂ it’s obvious that the derivative of a nonconstant polynomial f′ isn’t zero. But
suppose we considered the above as a polynomial in 𝔽3, i.e. modulo 3. Then the derivative is
zero. Oh, no!
We have to impose a condition that prevents something like this from happening.
Definition 52.4.1.For a field F, the characteristic of F is the smallest positive
integer p such that,
or zero if no such integer p exists.
Example 52.4.2 (Field characteristics) Old friends ℝ, ℚ, ℂ all have characteristic zero.
But 𝔽p, the integers modulo p, is a field of characteristic p.
Exercise 52.4.3.Let F be a field of characteristic p. Show that if p > 0 then p is a primenumber. (A proof is given next chapter.)
With the assumption of characteristic zero, our earlier proof works.
Lemma 52.4.4 (Separability in characteristic zero) Any irreducible polynomial in a
characteristic zero field is separable.
Unfortunately, this lemma is false if the “characteristic zero” condition is dropped.
Remark 52.4.5 — The reason it’s called separable is (I think) this picture: I have a
polynomial and I want to break it into irreducible parts. Normally, if I have a double
root in a polynomial, that means it’s not irreducible. But in characteristic p > 0 this
fails. So inseparable polynomials are strange when you think about them: somehow you
have double roots that can’t be separated from each other.
We can get this to work for any field extension in which separability is not an issue.
Definition 52.4.6.A separable extensionK∕F is one in which every irreducible
polynomial in F is separable (for example, if F has characteristic zero). A field F is
perfect if any finite field extension K∕F is separable.
In fact, as we see in the next chapter:
Theorem 52.4.7 (Finite fields are perfect) Suppose F is a field with finitely many
elements. Then it is perfect.
Thus, we will almost never have to worry about separability since every field we see in the
Napkin is either finite or characteristic 0. So the inclusion of the word “separable” is mostly a
formality.
Proceeding onwards, we obtain
Theorem 52.4.8 (The n embeddings of any separable extension) Let K∕F be a separable
extension of degree n and let F be an algebraic closure of F. Then there are exactly n
field homomorphisms KF, say σ1,…,σn, which fix F.
In any case, this lets us define the trace for any separable normal extension.
Definition 52.4.9.Let K∕F be a separable extension of degree n, and let σ1, …, σn be
the n embeddings into an algebraic closure of F. Then we define
When F = ℚ and the algebraic closure is ℂ, this coincides with our earlier definition!
52.5Automorphism groups and Galois extensions
Prototypical example for this section:ℚ() is Galois but ℚ() is not.
We now want to get back at the idea we stated at the beginning of this section that ℚ()
is deficient in a way that ℚ() is not.
First, we define the “internal” automorphisms.
Definition 52.5.1.Suppose K∕F is a finite extension. Then Aut(K∕F) is the set of
field isomorphisms σ : K → K which fix F. In symbols
This is a group under function composition!
Note that this time, we have a condition that F is fixed by σ. (This was not there before when
we considered F = ℚ, because we got it for free.)
Example 52.5.2 (Old examples of automorphism groups) Reprising the example at the
beginning of the chapter in the new notation, we have:
(a)
Aut(ℚ(i)∕ℚ)ℤ∕2ℤ, with elements zz and zz.
(b)
Aut(ℚ()∕ℚ)ℤ∕2ℤ in the same way.
(c)
Aut(ℚ()∕ℚ) is the trivial group, with only the identity embedding!
Example 52.5.3 (Automorphism group of ℚ(,)) Here’s a new example: let K =
ℚ(,). It turns out that Aut(K∕ℚ) = {1,σ,τ,στ}, where
In other words, Aut(K∕ℚ) is the Klein Four Group.
First, let’s repeat the proof of the observation that these embeddings shuffle around roots
(akin to the first observation in the introduction):
Lemma 52.5.4 (Root shuffling in Aut(K∕F)) Let f ∈ F[x], suppose K∕F is a finite
extension, and assume α ∈ K is a root of f. Then for any σ ∈ Aut(K∕F), σ(α) is also
a root of f.
Proof.Let f(x) = cnxn + cn−1xn−1 + + c0, where ci∈ F. Thus,
In particular, taking f to be the minimal polynomial of α we deduce
An embedding σ ∈ Aut(K∕F) sends an α ∈ K to one of its various Galois conjugates(over F).
Next, let’s look again at the “deficiency” of certain fields. Look at K = ℚ(). So, again
K∕ℚ is deficient for two reasons. First, while there are three maps ℚ()ℂ, only one of
them lives in Aut(K∕ℚ), namely the identity. In other words, is too small.
Secondly, K is missing some Galois conjugates (ω and ω2).
The way to capture the fact that there are missing Galois conjugates is the notion of a
splitting field.
Definition 52.5.5.Let F be a field and p(x) ∈ F[x] a polynomial of degree n. Then
p(x) has roots α1,…,αn in an algebraic closure of F. The splitting field of F is defined
as F(α1,…,αn).
In other words, the splitting field is the smallest field in which p(x) splits.
Example 52.5.6 (Examples of splitting fields)
(a)
The splitting field of x2− 5 over ℚ is ℚ(). This is a degree 2 extension.
(b)
The splitting field of x2 +x+1 over ℚ is ℚ(ω), where ω is a cube root of unity. This
is a degree 3 extension.
(c)
The splitting field of x2 + 3x + 2 = (x + 1)(x + 2) is just ℚ! There’s nothing to do.
Example 52.5.7 (Splitting fields: a cautionary tale) The splitting field of x3−2 over ℚ
is in fact
and not just ℚ()! One must really adjoin all the roots, and it’s not necessarily the
case that these roots will generate each other.
To be clear:
For x2− 5, we adjoin and this will automatically include −.
For x2 + x + 1, we adjoin ω and get the other root ω2 for free.
But for x3− 2, if we adjoin , we do NOT get ω and ω2 for free. Indeed,
ℚ() ⊂ ℝ!
Note that in particular, the splitting field of x3− 2 over ℚ is degree six, not just degree
three.
In general, the splitting field of a polynomial can be an extension of degree up ton!. The reason is that if p(x) has n roots and none of them are “related” to each other, then
any permutation of the roots will work.
Now, we obtain:
Theorem 52.5.8 (Galois extensions are splitting) For finite extensions K∕F,
divides [K : F], with equality if and only if K is the splitting field of some separable
polynomial with coefficients in F.
The proof of this is deferred to an optional section at the end of the chapter. If K∕F is a
finite extension and = [K : F], we say the extension K∕F is Galois. In that
case, we denote Aut(K∕F) by Gal(K∕F) instead and call this the Galois group of
K∕F.
Example 52.5.9 (Examples and non-examples of Galois extensions)
(a)
The extension ℚ()∕ℚ is Galois, since it’s the splitting field of x2− 2 over ℚ. The
Galois group has order two, ±.
(b)
The extension ℚ(,)∕ℚ is Galois, since it’s the splitting field of (x2− 5)2− 6
over ℚ. As discussed before, the Galois group is ℤ∕2ℤ × ℤ∕2ℤ.
(c)
The extension ℚ()∕ℚ is not Galois.
To explore ℚ() one last time:
Example 52.5.10 (Galois closures, and the automorphism group of ℚ(,ω)) Let’s return to the field K = ℚ(,ω), which is a field with [K : ℚ] = 6. Consider the
two automorphisms:
Notice that σ3 = τ2 = id. From this one can see that the automorphism group of K
must have order 6 (it certainly has order ≤ 6; now use Lagrange’s theorem). So, K∕ℚ is
Galois! Actually one can check explicitly that
is the symmetric group on 3 elements, with order 3! = 6.
This example illustrates the fact that given a non-Galois field extension, one can “add in”
missing conjugates to make it Galois. This is called taking a Galois closure.
52.6Fundamental theorem of Galois theory
After all this stuff about Galois Theory, I might as well tell you the fundamental theorem,
though I won’t prove it. Basically, it says that if K∕F is Galois with Galois group G,
then:
Subgroups of G correspond exactly to fields E with F ⊆ E ⊆ K.
To tell you how the bijection goes, I have to define a fixed field.
Definition 52.6.1.Let K be a field and H a subgroup of Aut(K∕F). We define the
fixed field of H, denoted KH, as
Question 52.6.2.Verify quickly that KHis actually a field.
Now let’s look at examples again. Consider K = ℚ(,), where
is the Klein four group (where σ() = − but σ() = ; τ goes the other
way).
Question 52.6.3.Let H = {id,σ}. What is KH?
In that case, the diagram of fields between ℚ and K matches exactly with the subgroups of
G, as follows:
We see that subgroups correspond to fixed fields. That, and much more, holds in general.
Theorem 52.6.4 (Fundamental theorem of Galois theory) Let K∕F be a Galois extension with
Galois group G = Gal(K∕F).
(a)
There is a bijection between field towers F ⊆ E ⊆ K and subgroups H ⊆ G:
The bijection sends H to its fixed field KH, and hence is inclusion reversing.
(b)
Under this bijection, we have [K : E] = and [E : F] = [G : H].
(c)
K∕E is always Galois, and its Galois group is Gal(K∕E) = H.
(d)
E∕F is Galois if and only if H is normal in G. If so, Gal(E∕F) = G∕H.
Exercise 52.6.5.Suppose we apply this theorem for
Verify that the fact E = ℚ() is not Galois corresponds to the fact that S3does not havenormal subgroups of order 2.
52.7A few harder problems to think about
Problem 52A⋆ (Galois group of the cyclotomic field).Let p be an odd rational prime
and ζp a primitive pth root of unity. Let K = ℚ(ζp). Show that
Problem 52B (Greek constructions).Prove that the three Greek constructions
(a)
doubling the cube,
(b)
squaring the circle, and
(c)
trisecting an angle
are all impossible. (Assume π is transcendental.)
Problem 52C (China Hong Kong Math Olympiad).Prove that
there are no rational numbers p, q, r satisfying
Problem 52D.Show that the only automorphism of ℝ is the identity. Hence Aut(ℝ) is the
trivial group.
Problem 52E (Artin’s primitive element theorem).Let K be a number
field. Show that Kℚ(γ) for some γ.
52.8(Optional) Proof that Galois extensions are splitting
We prove ?? . First, we extract a useful fragment from the fundamental theorem.
Theorem 52.8.1 (Fixed field theorem) Let K be a field and G a subgroup of Aut(K).
Then [K : KG] = .
The inequality itself is not difficult:
Exercise 52.8.2.Show that [K : F] ≥|Aut(K∕F)|, and that equality holds if and only if theset of elements fixed by all σ ∈Aut(K∕F) is exactly F. (Use?? .)
The equality case is trickier.
The easier direction is when K is a splitting field. Assume K = F(α1,…,αn) is the splitting
field of some separable polynomial p ∈ F[x] with n distinct roots α1,…,αn. Adjoin them one by
one:
(Does this diagram look familiar?) Every map K → K which fixes F corresponds to an above
commutative diagram. As before, there are exactly [F(α1) : F] ways to pick τ1. (You need the
fact that the minimal polynomial p1 of α1 is separable for this: there need to be exactly
deg p1 = [F(α1) : F] distinct roots to nail p1 into.) Similarly, given a choice of τ1, there are
[F(α1,α2) : F(α1)] ways to pick τ2. Multiplying these all together gives the desired
[K : F].
Now assume K∕F is Galois. First, we state:
Lemma 52.8.3 Let K∕F be Galois, and p ∈ F[x] irreducible. If any root of p (in F) lies
in K, then all of them do, and in fact p is separable.
Proof.Let α ∈ K be the prescribed root. Consider the set
(Note that α ∈ S since Gal(K∕F) ∋ id.) By construction, any τ ∈ Gal(K∕F) fixes S.
So if we construct
then by Vieta’s Formulas, we find that all the coefficients of p are fixed by elements
of σ. By the equality case we specified in the exercise, it follows that p has coefficients
in F! (This is where we use the condition.) Also, by ?? , p divides p.
Yet p was irreducible, so it is the minimal polynomial of α in F[x], and therefore we
must have that p divides p. Hence p = p. Since p was built to be separable, so is p. □
Now we’re basically done – pick a basis ω1, …, ωn of K∕F, and let pi be their minimal
polynomials; by the above, we don’t get any roots outside K. Consider P = p1…pn, removing
any repeated factors. The roots of P are ω1, …, ωn and some other guys in K. So K is the
splitting field of P.
53Finite fields
In this short chapter, we classify all fields with finitely many elements and compute the Galois
groups. Nothing in here is very hard, and so most of the proofs are just sketches; if you like,
you should check the details yourself.
The whole point of this chapter is to prove:
A finite field F must have order pn, with p prime and n an integer.
In this case, F has characteristic p.
All such fields are isomorphic, so it’s customary to use the notation 𝔽pn for “the”
finite field of order pn if we only care up to isomorphism.
The extension F∕𝔽p is Galois, and Gal(F∕𝔽p) is a cyclic group of order n. The
generator is the automorphism
If you’re in a hurry you can just remember these results and skip to the next chapter.
53.1Example of a finite field
Before diving in, we give some examples.
Recall that the characteristic of a field F is the smallest positive integer p such
that
or 0 if no such integer p exists.
Example 53.1.1 (Base field) Let 𝔽p denote the field of integers modulo p. This is a field
with p elements, with characteristic p.
Example 53.1.2 (The finite field of nine elements) Let
We can think of its elements as
Since (3) is prime in ℤ[i], the ring of integers of ℚ(i), we see F is a field with 32 = 9
elements inside it. Note that, although this field has 9 elements, every element x has the
property that
In particular, F has characteristic 3.
53.2Finite fields have prime power order
Lemma 53.2.1 If the characteristic of a field F isn’t zero, it must be a prime number.
Proof.Assume not, so n = ab for a,b < n. Then let
and
Then AB = 0, contradicting the fact that F is a field. □
We like fields of characteristic zero, but unfortunately for finite fields we are doomed to have
nonzero characteristic.
Lemma 53.2.2 (Finite fields have prime power orders) Let F be a finite field. Then
(a)
Its characteristic is nonzero, and hence some prime p.
(b)
The field F is a finite extension of 𝔽p, and in particular it is an 𝔽p-vector space.
(c)
We have = pn for some prime p, integer n.
Proof.Very briefly, since this is easy:
(a)
Apply Lagrange’s theorem (or pigeonhole principle!) to (F,+) to get the
characteristic isn’t zero.
(b)
The additive subgroup of (F,+) generated by 1F is an isomorphic copy of 𝔽p.
(c)
Since it’s a field extension, F is a finite-dimensional vector space over 𝔽p, with some
basis e1,…,en. It follows that there are pn elements of F. □
Remark 53.2.3 — An amusing alternate proof of (c) by contradiction: if a prime q≠p
divides , then by Cauchy’s theorem (?? ) on (F,+) there’s a (nonzero) element x of
order q. Evidently
then, but x≠0, and hence the characteristic of F also divides q, which is impossible.
An important point in the above proof is that
Lemma 53.2.4 (Finite fields are field extensions of 𝔽p) If = pn is a finite field, then
there is an isomorphic copy of 𝔽p sitting inside F. Thus F is a field extension of 𝔽p.
We want to refer a lot to this copy of 𝔽p, so in what follows:
Abuse of Notation 53.2.5.Every integer n can be identified as an element of F,
namely
Note that (as expected) this depends only on n (mod p).
This notation makes it easier to think about statements like the following.
Theorem 53.2.6 (Freshman’s dream) For any a,b ∈ F we have
Proof.Use the Binomial theorem, and the fact that is divisible by p for 0 < i < p. □
Exercise 53.2.7.Convince yourself that this proof works.
53.3All finite fields are isomorphic
We next proceed to prove “Fermat’s little theorem”:
Theorem 53.3.1 (Fermat’s little theorem in finite fields) Let F be a finite field of order pn.
Then every element x ∈ F satisfies
Proof.If x = 0 it’s true; otherwise, use Lagrange’s theorem on the abelian group (F,×)
to get xpn−1 = 1F. □
We can now prove the following result, which is the “main surprise” about finite fields: that
there is a unique one up to isomorphism for each size.
Theorem 53.3.2 (Complete classification of finite fields) A field F is a finite field with pn
elements if and only if it is a splitting field of xpn− x over 𝔽p.
Proof.By “Fermat’s little theorem”, all the elements of F satisfy this polynomial. So we
just have to show that the roots of this polynomial are distinct (i.e. that it is separable).
To do this, we use the derivative trick again: the derivative of this polynomial is
which has no roots at all, so the polynomial cannot have any double roots. □
Definition 53.3.3.For this reason, it’s customary to denote the field with pn elements
by 𝔽pn.
Note that the polynomial xpn− x (mod p) is far from irreducible, but the computation
above shows that it’s separable.
Example 53.3.4 (The finite field of order nine again) The polynomial x9− x is
separable modulo 3 and has factorization
So if F has order 9, then we intuitively expect it to be the field generated by adjoining
all the roots: 0, 1, 2, as well as ±i, 1 ±i, 2 ±i. Indeed, that’s the example we had at the
beginning of this chapter.
(Here i denotes an element of 𝔽9 satisfying i2 = −1. The notation is deliberately similar
to the usual imaginary unit.)
53.4The Galois theory of finite fields
Retain the notation 𝔽pn now (instead of F like before). By the above theorem, it’s the
splitting field of a separable polynomial, hence we know that 𝔽pn∕𝔽p is a Galois extension. We
would like to find the Galois group.
In fact, we are very lucky: it is cyclic. First, we exhibit one such element σp∈ Gal(𝔽pn∕𝔽p):
Theorem 53.4.1 (The pth power automorphism) The map σp : 𝔽pn→ 𝔽pn defined by
is an automorphism, and moreover fixes 𝔽p.
Proof.It’s a homomorphism since it fixes 1, respects multiplication, and respects
addition.
Question 53.4.2.Why does it respect addition?
Next, we claim that it is injective. To see this, note that
Here we have again used the Freshman’s Dream. Since 𝔽pn is finite, this injective map is
automatically bijective. The fact that it fixes 𝔽p is Fermat’s little theorem. □
Now we’re done:
Theorem 53.4.3 (Galois group of the extension 𝔽pn∕𝔽p) We have Gal(𝔽pn∕𝔽p)ℤ∕nℤ
with generator σp.
Proof.Since [𝔽pn : 𝔽p] = n, the Galois group G has order n. So we just need to show
σp∈ G has order n.
Note that σp applied k times gives xxpk. Hence, σp applied n times is the identity,
as all elements of 𝔽pn satisfy xpn = x. But if k < n, then σp applied k times cannot be
the identity or xpk− x would have too many roots. □
We can see an example of this again with the finite field of order 9.
Example 53.4.4 (Galois group of finite field of order 9) Let 𝔽9 be the finite field of order 9,
and represent it concretely by 𝔽9 = ℤ[i]∕(3). Let σ3 : 𝔽9→ 𝔽9 be xx3. We can witness the
fate of all nine elements:
(As claimed, 0, 1, 2 are the fixed points, so I haven’t drawn arrows for them.) As predicted,
the Galois group has order two:
This concludes the proof of all results stated at the beginning of this chapter.
53.5A few harder problems to think about
Problem 53A† (HMMT 2017).What is the period of the Fibonacci
sequence modulo 127?
54Ramification theory
We’re very interested in how rational primes p factor in a bigger number field K. Some
examples of this behavior: in ℤ[i] (which is a UFD!), we have factorizations
(2)
= (1 + i)2
(3)
= (3)
(5)
= (2 + i)(2 − i).
In this chapter we’ll learn more about how primes break down when they’re thrown into
bigger number fields. Using weapons from Galois Theory, this will culminate in a proof of
Quadratic Reciprocity.
54.1Ramified / inert / split primes
Prototypical example for this section:In ℤ[i], 2 is ramified, 3 is inert, and 5 splits.
Let p be a rational prime, and toss it into 𝒪K. Thus we get a factorization into prime
ideals
We say that each 𝔭i is above (p).1
Pictorially, you might draw this as follows:
Some names for various behavior that can happen:
We say p is ramified if ei> 1 for some i. For example 2 is ramified in ℤ[i].
We say p is inert if g = 1 and e1 = 1; i.e. (p) remains prime. For example 3 is
inert in ℤ[i].
We say p is split if g > 1. For example 5 is split in ℤ[i].
Question 54.1.1.More generally, for a prime p in ℤ[i]:
p is ramified exactly when p = 2.
p is inert exactly when p ≡ 3 (mod4).
p is split exactly when p ≡ 1 (mod4).
Prove this.
54.2Primes ramify if and only if they divide ΔK
The most unusual case is ramification: Just like we don’t expect a randomly selected
polynomial to have a double root, we don’t expect a randomly selected prime to be ramified.
In fact, the key to understanding ramification is the discriminant.
For the sake of discussion, let’s suppose that K is monogenic, 𝒪K = ℤ[𝜃], where 𝜃 has
minimal polynomial f. Let p be a rational prime we’d like to factor. If f factors as f1e1…fgeg,
then we know that the prime factorization of (p) is given by
In particular, p ramifies exactly when f has a double root mod p! To detect whether this
happens, we look at the polynomial discriminant of f, namely
and see whether it is zero mod p – thus p ramifies if and only if this is true.
It turns out that the naïve generalization to any number field works if we replace Δ(f) by
just the discriminant ΔK of K; (these are the same for monogenic 𝒪K by ?? ). That
is,
Theorem 54.2.1 (Discriminant detects ramification) Let p be a rational prime and K a
number field. Then p is ramified if and only if p divides ΔK.
Example 54.2.2 (Ramification in the Gaussian integers) Let K = ℚ(i) so 𝒪K = ℤ[i]
and ΔK = 4. As predicted, the only prime ramifying in ℤ[i] is 2, the only prime factor
of ΔK.
In particular, only finitely many primes ramify.
54.3Inertial degrees
Prototypical example for this section: (7) has inertial degree 2 in ℤ[i] and (2 + i) has inertialdegree 1 in ℤ[i].
Recall that we were able to define an ideal norm N(𝔞) = measuring how “roomy”
the ideal 𝔞 is. For example, (5) has ideal norm 52 = 25 in ℤ[i], since
has 52 = 25 elements.
Now, let’s look at
in 𝒪K, where K has degree n. Taking the ideal norms of both sides, we have
that
We conclude that 𝔭i = pfi for some integer fi≥ 1, and moreover that
Definition 54.3.1.We say fi is the inertial degree of 𝔭i, and ei is the ramificationindex.
Example 54.3.2 (Examples of inertial degrees) Work in ℤ[i], which is degree 2. The inertial
degree detects how “spacy” the given 𝔭 is when interpreted in 𝒪K.
(a)
The prime 7 ⋅ ℤ[i] has inertial degree 2. Indeed, ℤ[i]∕(7) has 72 = 49 elements, those
of the form a + bi for a, b modulo 7. It gives “two degrees” of space.
(b)
Let (5) = (2+i)(2−i). The inertial degrees of (2+i) and (2−i) are both 1. Indeed,
ℤ[i]∕(2+i) only gives “one degree” of space, since each of its elements can be viewed
as integers modulo 5, and there are only 51 = 5 elements.
If you understand this, it should be intuitively clear why the sum of eifi should equal
n.
54.4The magic of Galois extensions
OK, that’s all fine and well. But something really magical happens when we add the
additional hypothesis that K∕ℚ is Galois: all the inertial degrees and ramification degrees are
equal. We set about proving this.
Let K∕ℚ be Galois with G = Gal(K∕ℚ). Note that if 𝔭 ⊆𝒪K is a prime above p, then the
image σimg(𝔭) is also prime for any σ ∈ G (since σ is an automorphism!). Moreover, since
p ∈𝔭 and σ fixes ℚ, we know that p ∈ σimg(𝔭) as well.
Thus, by the pointwise mapping, the Galois group acts on the prime ideals above arational prime p. Picture:
The notation σimg(𝔭) is hideous in this context, since we’re really thinking of σ as just
doing a group action, and so we give the shorthand:
Abuse of Notation 54.4.1.Let σ𝔭 be shorthand for σimg(𝔭).
Since the σ’s are all bijections (they are automorphisms!), it should come as no surprise that
the prime ideals which are in the same orbit are closely related. But miraculously, it turns out
there is only one orbit!
Theorem 54.4.2 (Galois group acts transitively) Let K∕ℚ be Galois with G = Gal(K∕ℚ).
Let {𝔭i} be the set of distinct prime ideals in the factorization of p ⋅𝒪K (in 𝒪K).
Then G acts transitively on the 𝔭i: for every i and j, we can find σ such that σ𝔭i = 𝔭j.
Proof.Fairly slick. Suppose for contradiction that no σ ∈ G sends 𝔭1 to 𝔭2, say. By the
Chinese remainder theorem, we can find an x ∈𝒪K such that
x
≡ 0 (mod 𝔭1)
x
≡ 1 (mod 𝔭i) for i ≥ 2
Then, compute the norm
Each σ(x) is in K because K∕ℚ is Galois!
Since NK∕ℚ(x) is an integer and divisible by 𝔭1, we should have that NK∕ℚ(x) is divisible by
p. Thus it should be divisible by 𝔭2 as well. But by the way we selected x, we have xσ−1𝔭2
for every σ ∈ G! So σ(x)𝔭2 for any σ, which is a contradiction. □
Theorem 54.4.3 (Inertial degree and ramification indices are all equal) Assume K∕ℚ is
Galois. Then for any rational prime p we have
for some e, where the 𝔭i are distinct prime ideals with the same inertial degree f. Hence
Proof.To see that the inertial degrees are equal, note that each σ induces an isomorphism
Because the action is transitive, all fi are equal.
Exercise 54.4.4.Using the fact that σ ∈Gal(K∕ℚ), show that
So for every σ, we have that p⋅𝒪K = ∏𝔭iei = ∏(σ𝔭i)ei. Since the action is transitive, all ei
are equal. □
Let’s see an illustration of this.
Example 54.4.5 (Factoring 5 in a Galois/non-Galois extension) Let p = 5 be a
prime.
(a)
Let E = ℚ(). One can show that 𝒪E = ℤ[], so we use the Factoring Algorithm
on the minimal polynomial x3− 2. Since x3− 2 ≡ (x − 3)(x2 + 3x + 9) (mod 5) is
the irreducible factorization, we have that
which have inertial degrees 1 and 2, respectively. The fact that this is not uniform
reflects that E is not Galois.
(b)
Now let K = ℚ(,ω), which is the splitting field of x3−2 over ℚ; now K is Galois.
It turns out that
(this takes a lot of work to obtain, so we won’t do it here). Modulo 5 this has an
irreducible factorization (x2 + x + 2)(x2 + 3x + 3)(x2 + 4x + 1) (mod 5), so by the
Factorization Algorithm,
This time all inertial degrees are 2, as the theorem predicts for K Galois.
54.5(Optional) Decomposition and inertia groups
Let p be a rational prime. Thus
and all the 𝔭i have inertial degree f. Let 𝔭 denote a choice of the 𝔭i.
We can look at both the fields 𝒪K∕𝔭 and ℤ∕p = 𝔽p. Naturally, since 𝒪K∕𝔭 is
a finite field we can view it as a field extension of 𝔽p. So we can get the diagram
At the far right we have finite field extensions, which we know are really well behaved. So we
ask:
How are Galand Gal(K∕ℚ) related?
Absurdly enough, there is an explicit answer: it’s just the stabilizer of 𝔭, at least when p isunramified.
Definition 54.5.1.Let D𝔭⊆ Gal(K∕ℚ) be the stabilizer of 𝔭, that is
We say D𝔭 is the decomposition group of 𝔭.
Then, every σ ∈ D𝔭 induces an automorphism of 𝒪K∕𝔭 by
So there’s a natural map
by declaring 𝜃(σ) to just be “σ (mod 𝔭)”. The fact that σ ∈ D𝔭 (i.e. σ fixes 𝔭) ensures this
map is well-defined.
Theorem 54.5.2 (Decomposition group and Galois group) Define 𝜃 as above. Then
𝜃 is surjective, and
its kernel is a group of order e, the ramification index.
In particular, if p is unramified then D𝔭Gal.
(The proof is not hard, but a bit lengthy and in my opinion not very enlightening.)
If p is unramified, then taking modulo 𝔭 gives D𝔭Gal.
But we know exactly what Gal is! We already have 𝒪K∕𝔭𝔽pf, and the
Galois group is
So
as well.
Let’s now go back to
The kernel of 𝜃 is called the inertia group and denoted I𝔭⊆ D𝔭; it has order
e.
This gives us a pretty cool sequence of subgroups {1}⊆ I ⊆ D ⊆ G where G is the Galois
group (I’m dropping the 𝔭-subscripts now). Let’s look at the corresponding fixed fields via the
Fundamental theorem of Galois theory. Picture:
Something curious happens:
When (p) is lifted into KD it splits completely into g unramified primes. Each of
these has inertial degree 1.
When the primes in KD are lifted to KI, they remain inert, and now have inertial
degree f.
When then lifted to K, they ramify with exponent e (but don’t split at all).
Picture: In other words, the process of going from 1 to efg can be very nicely broken into the
three steps above. To draw this in the picture, we get
In any case, in the “typical” case that there is no ramification, we just have KI = K.
54.6Tangential remark: more general Galois extensions
All the discussion about Galois extensions carries over if we replace K∕ℚ by some different
Galois extension K∕F. Instead of a rational prime p breaking down in 𝒪K, we would have a
prime ideal 𝔭 of F breaking down as
in 𝒪L and then all results hold verbatim. (The 𝔓i are primes in L above 𝔭.) Instead of 𝔽p
we would have 𝒪F∕𝔭.
The reason I choose to work with F = ℚ is that capital Gothic P’s (𝔓) look really
terrifying.
54.7A few harder problems to think about
Problem 54A†.Prove that no rational prime p can remain inert in K = ℚ(,ω), the
splitting field of x3− 2. How does this generalize?
55The Frobenius element
Throughout this chapter K∕ℚ is a Galois extension with Galois group G, p is an unramified
rational prime in K, and 𝔭 is a prime above it. Picture:
If p is unramified, then one can show there is a unique σ ∈ Gal(L∕K) such that σ(α) ≡ αp
(mod 𝔭) for every prime p.
55.1Frobenius elements
Prototypical example for this section: Frob𝔭in ℤ[i] depends on p (mod 4).
Here is the theorem statement again:
Theorem 55.1.1 (The Frobenius element) Assume K∕ℚ is Galois with Galois group G.
Let p be a rational prime unramified in K, and 𝔭 a prime above it. There is a unique
element Frob𝔭∈ G with the property that
It is called the Frobenius element at 𝔭, and has order f.
The uniqueness part is pretty important: it allows us to show that a given σ ∈ Gal(L∕K)
is the Frobenius element by just observing that it satisfies the above functional
equation.
Let’s see an example of this:
Example 55.1.2 (Frobenius elements of the Gaussian integers) Let’s actually compute
some Frobenius elements for K = ℚ(i), which has 𝒪K = ℤ[i]. This is a Galois extension,
with G = (ℤ∕2ℤ)×, corresponding to the identity and complex conjugation.
If p is an odd prime with 𝔭 above it, then Frob𝔭 is the unique element such that
in ℤ[i]. In particular,
From this we see that Frob𝔭 is the identity when p ≡ 1 (mod 4) and Frob𝔭 is complex
conjugation when p ≡ 3 (mod 4).
Note that we really only needed to compute Frob𝔭 on i. If this seems too good to be true, a
philosophical reason is “freshman’s dream” where (x + y)p≡ xp + yp (mod p) (and hence mod
𝔭). So if σ satisfies the functional equation on generators, it satisfies the functional equation
everywhere.
We also have an important lemma:
Lemma 55.1.3 (Order of the Frobenius element) Let Frob𝔭 be a Frobenius element from
an extension K∕ℚ. Then the order of 𝔭 is equal to the inertial degree f𝔭. In particular,
(p) splits completely in 𝒪K if and only if Frob𝔭 = id.
Exercise 55.1.4.Prove this lemma as by using the fact that 𝒪K∕𝔭 is the finite field of orderf𝔭, and the Frobenius element is just xxpon this field.
Let us now prove the main theorem. This will only make sense in the context of
decomposition groups, so readers which skipped that part should omit this proof.
Proof of existence of Frobenius element. The entire theorem is just a rephrasing of the fact
that the map 𝜃 defined in the last section is an isomorphism when p is unramified. Picture:
In here we can restrict our attention to D𝔭 since we need to have σ(α) ≡ 0 (mod 𝔭) when α ≡ 0
(mod 𝔭). Thus we have the isomorphism
But we already know Gal, according to the string of isomorphisms
So the unique such element is the pre-image of T under 𝜃. □
55.2Conjugacy classes
Now suppose 𝔭1 and 𝔭2 are two primes above an unramified rational prime p. Then we can
define Frob𝔭1 and Frob𝔭2. Since the Galois group acts transitively, we can select σ ∈ Gal(K∕ℚ)
be such that
We claim that
Note that this is an equation in G.
Question 55.2.1.Prove this.
More generally, for a given unramified rational prime p, we obtain:
Theorem 55.2.2 (Conjugacy classes in Galois groups) The set
is one of the conjugacy classes of G.
Proof.We’ve used the fact that G = Gal(K∕ℚ) is transitive to show that Frob𝔭1 and
Frob𝔭2 are conjugate if they both lie above p; hence it’s contained in some conjugacy
class. So it remains to check that for any 𝔭, σ, we have σ∘Frob𝔭∘σ−1 = Frob𝔭′ for some
𝔭′. For this, just take 𝔭′ = σ𝔭. Hence the set is indeed a conjugacy class. □
In summary,
Frob𝔭is determined up to conjugation by the prime p from which 𝔭 arises.
So even though the Gothic letters look scary, the content of Frob𝔭 really just comes from the
more friendly-looking rational prime p.
Example 55.2.3 (Frobenius elements in ℚ(,ω)) With those remarks, here is a more
involved example of a Frobenius map. Let K = ℚ(,ω) be the splitting field of
Thus K∕ℚ is Galois. We’ve seen in an earlier example that
Let’s consider the prime 5 which factors (trust me here) as
Note that all the prime ideals have inertial degree 2. Thus Frob𝔭i will have order 2 for
each i.
Note that
In this S3 there are 3 elements of order three: fixing one root and swapping the other
two. These correspond to each of Frob𝔭1, Frob𝔭2, Frob𝔭3.
In conclusion, the conjugacy class associated to (5) is the cycle
type (∙)(∙∙) in S3.
55.3Chebotarev density theorem
Natural question: can we represent every conjugacy class in this way? In other words, is every
element of G equal to Frob𝔭 for some 𝔭?
Miraculously, not only is the answer “yes”, but in fact it does so in the nicest way possible:
the Frob𝔭’s are “equally distributed” when we pick a random 𝔭.
Theorem 55.3.1 (Chebotarev density theorem over ℚ) Let C be a conjugacy class of G =
Gal(K∕ℚ). The density of (unramified) primes p such that {Frob𝔭∣𝔭 above p} = C is
exactly ∕. In particular, for any σ ∈ G there are infinitely many rational primes
p with 𝔭 above p so that Frob𝔭 = σ.
By density, I mean that the proportion of primes p ≤ x that work approaches as x →∞.
Note that I’m throwing out the primes that ramify in K. This is no issue, since the
only primes that ramify are those dividing ΔK, of which there are only finitely
many.
In other words, if I pick a random prime p and look at the resulting conjugacy class, it’s a
lot like throwing a dart at G: the probability of hitting any conjugacy class depends just on
the size of the class.
Remark 55.3.2 — Happily, this theorem (and preceding discussion) also works if we
replace K∕ℚ with any Galois extension K∕F; in that case we replace “𝔭 over p” with “𝔓
over 𝔭”. In that case, we use N(𝔭) ≤ x rather than p ≤ x as the way to define density.
55.4Example: Frobenius elements of cyclotomic fields
Let q be a prime, and consider L = ℚ(ζq), with ζq a primitive qth root of unity. You should
recall from various starred problems that
ΔL = ±qq−2,
𝒪L = ℤ[ζq], and
The map
is an automorphism of L whenever gcd(n,q) = 1, and depends only on n (mod q).
In other words, the automorphisms of L∕ℚ just shuffle around the qth roots of
unity. In fact the Galois group consists exactly of the elements {σn}, namely
As a group,
This is surprisingly nice, because elements of Gal(L∕ℚ) look a lot like Frobenius elementsalready. Specifically:
Lemma 55.4.1 (Cyclotomic Frobenius elements) In the cyclotomic setting L = ℚ(ζq), let
p be a rational unramified prime and 𝔭 above it. Then
Proof.Observe that σp satisfies the functional equation (check on generators). Done by
uniqueness. □
Question 55.4.2.Conclude that a rational prime p splits completely in 𝒪Lif and only if p ≡ 1(modm).
55.5Frobenius elements behave well with restriction
Let L∕ℚ and K∕ℚ be Galois extensions, and consider the setup
Here 𝔭 is above (p) and 𝔓 is above 𝔭. We may define
and want to know how these are related.
Theorem 55.5.1 (Restrictions of Frobenius elements) Assume L∕ℚ and K∕ℚ are both
Galois. Let 𝔓 and 𝔭 be unramified as above. Then Frob𝔓↾K = Frob𝔭, i.e. for every
α ∈ K,
Proof.We know
from the definition.
Question 55.5.2.Deduce that
(This is weaker than the previous statement in two ways!)
Thus Frob𝔓 restricted to 𝒪K satisfies the characterizing property of Frob𝔭. □
In short, the point of this section is that
Frobenius elements upstairs restrict to Frobenius elements downstairs.
55.6Application: Quadratic reciprocity
We now aim to prove:
Theorem 55.6.1 (Quadratic reciprocity) Let p and q be distinct odd primes. Then
(See, e.g. [?] for an exposition on quadratic reciprocity, if you’re not familiar with
it.)
55.6.iStep 1: Setup
For this proof, we first define
where ζq is a primitive qth root of unity. Then L∕ℚ is Galois, with Galois group
G.
Question 55.6.2.Show that G has a unique subgroup H of order two.
In fact, we can describe it exactly: viewing G(ℤ∕qℤ)×, we have
By the fundamental theorem of Galois Theory, there ought to be a degree 2 extension of ℚ
inside ℚ(ζq) (that is, a quadratic field). Call it ℚ(), for q∗ squarefree:
Exercise 55.6.3.Note that if a rational prime ℓ ramifies in K, then it ramifies in L. Use thisto show that
Together these determine the value of q∗.
(Actually, it is true in general ΔK divides ΔL in a tower L∕K∕ℚ.)
55.6.iiStep 2: Reformulation
Now we are going to prove:
Theorem 55.6.4 (Quadratic reciprocity, equivalent formulation) For distinct odd primes p,
q we have
Exercise 55.6.5.Using the fact that = (−1), show that this is equivalent to quadraticreciprocity as we know it.
We look at the rational prime p in ℤ. Either it splits into two in K or is inert; either way let
𝔭 be a prime factor in the resulting decomposition (so 𝔭 is either p ⋅𝒪K in the inert case, or
one of the primes in the split case). Then let 𝔓 be above 𝔭. It could possibly also split in K:
the picture looks like
Question 55.6.6.Why is p not ramified in either K or L?
55.6.iiiStep 3: Introducing the Frobenius
Now, we take the Frobenius
We claim that
To see this, note that Frob𝔓 is in H if and only if it acts as the identity on K. But Frob𝔓↾K
is Frob𝔭! So
Finally note that Frob𝔭 has order 1 if p splits (𝔭 has inertial degree 1) and order 2 if p is
inert. This completes the proof of the claim.
55.6.ivFinishing up
We already know by ?? that Frob𝔓 = σp∈ H if and only if p is a quadratic residue. On the
other hand,
Exercise 55.6.7.Show that p splits in 𝒪K= ℤ[(1 +)] if and only if = 1. (Use thefactoring algorithm. You need the fact that p≠2 here.)
In other words
This completes the proof.
55.7Frobenius elements control factorization
Prototypical example for this section: Frob𝔭controlled the splitting of p in the proof ofquadratic reciprocity; the same holds in general.
In the proof of quadratic reciprocity, we used the fact that Frobenius elements
behaved well with restriction in order to relate the splitting of p with properties of
Frob𝔭.
In fact, there is a much stronger statement for any intermediate field ℚ ⊆ E ⊆ K which
works even if E∕ℚ is not Galois. It relies on the notion of a factorization pattern. Here is how
it goes.
Set n = [E : ℚ], and let p be a rational prime unramified in K. Then p can be broken in E
as
with inertial degrees f1, …, fg: (these inertial degrees might be different since E∕ℚ isn’t
Galois). The numbers f1 + + fg = n form a partition of the number n. For
example, in the quadratic reciprocity proof we had n = 2, with possible partitions 1 + 1
(if p split) and 2 (if p was inert). We call this the factorization pattern of p in
E.
Next, we introduce a Frobenius Frob𝔓 above (p), all the way in K; this is an element of
G = Gal(K∕ℚ). Then let H be the group corresponding to the field E. Diagram:
Then Frob𝔓 induces a permutation of the n left cosets gH by left multiplication (after all,
Frob𝔓 is an element of G too!). Just as with any permutation, we may look at the
resulting cycle decomposition, which has a natural “cycle structure”: a partition of n.
The theorem is that these coincide:
Theorem 55.7.1 (Frobenius elements control decomposition) Let ℚ ⊆ E ⊆ K an
extension of number fields and assume K∕ℚ is Galois (though E∕ℚ need not be). Pick an
unramified rational prime p; let G = Gal(K∕ℚ) and H the corresponding intermediate
subgroup. Finally, let 𝔓 be a prime above p in K.
Then the factorization pattern of p in E is given by the cycle structure of Frob𝔓 acting
on the left cosets of H.
Often, we take E = K, in which case this is just asserting that the decomposition of the
prime p is controlled by a Frobenius element over it.
An important special case is when E = ℚ(α), because as we will see it is let us determine
how the minimal polynomial of α factors modulo p. To motivate this, let’s go back a few
chapters and think about the Factoring Algorithm.
Let α be an algebraic integer and f its minimal polynomial (of degree n). Set E = ℚ(α)
(which has degree n over ℚ). Suppose we’re lucky enough that 𝒪E = ℤ[α], i.e. that E
is monogenic. Then we know by the Factoring Algorithm, to factor any p in E,
all we have to do is factor f modulo p, since if f = f1e1…fgeg (mod p) then we
have
This gives us complete information about the ramification indices and inertial
degrees; the ei are the ramification indices, and deg fi are the inertial degrees (since
𝒪E∕𝔭i𝔽p[X]∕(fi(X))).
In particular, if p is unramified then all the ei are equal to 1, and we get
Once again we have a partition of n; we call this the factorization pattern of f modulo p.
So, to see the factorization pattern of an unramified p in 𝒪E, we just have to know the
factorization pattern of the f (mod p).
Turning this on its head, if we want to know the factorization pattern of f (mod p), we just
need to know how p decomposes. And it turns out these coincide even without the assumption
that E is monogenic.
Theorem 55.7.2 (Frobenius controls polynomial factorization) Let α be an algebraic integer
with minimal polynomial f, and let E = ℚ(α). Then for any prime p unramified in the
splitting field K of f, the following coincide:
(i)
The factorization pattern of p in E.
(ii)
The factorization pattern of f (mod p).
(iii)
The cycle structure associated to the action of Frob𝔓∈ Gal(K∕ℚ) on the roots of
f, where 𝔓 is above p in K.
Example 55.7.3 (Factoring x3− 2 (mod 5)) Let α = and f = x3− 2, so E = ℚ(). Set
p = 5 and let finally, let K = ℚ(,ω) be the splitting field. Setup:
The three claimed objects now all have shape 2 + 1:
(i)
By the Factoring Algorithm, we have (5) = (5,− 3)(5,9 + 3 + ).
(ii)
We have x3− 2 ≡ (x − 3)(x2 + 3x + 9) (mod 5).
(iii)
We saw before that Frob𝔓 = (∙)(∙∙).
Sketch of Proof. Letting n = deg f. Let H be the subgroup of G = Gal(K∕ℚ) corresponding
to E, so [G : E] = n. Pictorially, we have
We claim that (i), (ii), (iii) are all equivalent to
(iv) The pattern of the action of Frob𝔓 on the G∕H.
In other words we claim the cosets correspond to the n roots of f in K. Indeed H is just the set
of τ ∈ G such that τ(α) = α, so there’s a bijection between the roots and the cosets G∕H by
τHτ(α). Think of it this way: if G = Sn, and H = {τ : τ(1) = 1}, then G∕H has order
n!∕(n− 1)! = n and corresponds to the elements {1,…,n}. So there is a natural bijection from
(iii) to (iv).
The fact that (i) is in bijection to (iv) was the previous theorem, ?? . The correspondence
(i) ⇔ (ii) is a fact of Galois theory, so we omit the proof here. □
All this can be done in general with ℚ replaced by F; for example, in [?].
55.8Example application: IMO 2003 problem 6
As an example of the power we now have at our disposal, let’s prove:
Problem 6. Let p be a prime number. Prove that there
exists a prime number q such that for every integer n, the
number np− p is not divisible by q.
We will show, much more strongly, that there exist infinitely many primes q such that Xp−p is
irreducible modulo q.
Solution. Okay! First, we draw the tower of fields
where K is the splitting field of f(x) = xp− p. Let E = ℚ() for brevity and note it has
degree [E : ℚ] = p. Let G = Gal(K∕ℚ).
Question 55.8.1.Show that p divides the order of G. (Look at E.)
Hence by Cauchy’s theorem (?? , which is a purely group-theoretic fact) we can find a
σ ∈ G of order p. By Chebotarev, there exist infinitely many rational (unramified) primes q≠p
and primes 𝔔 ⊆𝒪K above q such that Frob𝔔 = σ. (Yes, that’s an uppercase Gothic Q.
Sorry.)
We claim that all these q work.
By ?? , the factorization of f (mod q) is controlled by the action of σ = Frob𝔔 on the roots
of f. But σ has prime order p in G! So all the lengths in the cycle structure have to divide p.
Thus the possible factorization patterns of f are
So we just need to rule out the p = 1 + + 1 case now: this only happens if f
breaks into linear factors mod q. Intuitively this edge case seems highly unlikely
(are we really so unlucky that f factors into linear factors when we want it to be
irreducible?). And indeed this is easy to see: this means that σ fixes all of the roots
of f in K, but that means σ fixes K altogether, and hence is the identity of G,
contradiction. □
Remark 55.8.2 — In fact K = ℚ(,ζp), and = p(p−1). With a little more group
theory, we can show that in fact the density of primes q that work is .
55.9A few harder problems to think about
Problem 55A.Show that for an odd prime p,
Problem 55B.Let f be a nonconstant polynomial with integer coefficients. Suppose
f (mod p) splits completely into linear factors for all sufficiently large primes p. Show
that f splits completely into linear factors.
Problem 55C† (Dirichlet’s theorem on arithmetic progressions).Let a and m be relatively
prime positive integers. Show that the density of primes p ≡ a (mod m) is exactly .
Problem 55D.Let n be an odd integer which is not a prime power. Show that the nth
cyclotomic polynomial is not irreducible modulo any rational prime.
Problem 55E (Putnam 2012 B6).Let p be an odd prime such that p ≡ 2
(mod 3). Let π be a permutation of 𝔽p by π(x) = x3 (mod p). Show that π is even if and only
if p ≡ 3 (mod 4).
56Bonus: A Bit on Artin Reciprocity
In this chapter, I’m going to state some big theorems of global class field theory
and use them to deduce the Kronecker-Weber plus Hilbert class fields. No proofs,
but hopefully still appreciable. For experts: this is global class field theory, without
ideles.
Here’s the executive summary: let K be a number field. Then all abelian extensions L∕K
can be understood using solely information intrinsic to K: namely, the ray class groups
(generalizing ideal class groups).
56.1Infinite primes
Prototypical example for this section:ℚ() has a complex infinite prime, ℚ() has tworeal infinite ones.
Let K be a number field of degree n and signature (r,s). We know what a prime ideal of
𝒪K is, but we now allow for the so-called infinite primes, which I’ll describe using the
embeddings.1
Recall there are n embeddings σ : K → ℂ, which consist of
r real embeddings where imσ ⊆ ℝ, and
s pairs of conjugate complex embeddings.
Hence r + 2s = n. The first class of embeddings form the real infinite primes, while the
complex infinite primes are the second type. We say K is totally real (resp totallycomplex) if all its infinite primes are real (resp complex).
Example 56.1.1 (Examples of infinite primes)
ℚ has a single real infinite prime. We often write it as ∞.
ℚ() has a single complex infinite prime, and no real infinite primes. Hence
totally complex.
ℚ() has two real infinite primes, and no complex infinite primes. Hence totally
real.
56.2Modular arithmetic with infinite primes
A modulus is a formal product
where the product runs over all primes, finite and infinite. (Here ν(𝔭) is a nonnegative
integer, of which only finitely many are nonzero.) We also require that
ν(𝔭) = 0 for any complex infinite prime 𝔭, and
ν(𝔭) ≤ 1 for any real infinite prime 𝔭.
Obviously, every 𝔪 can be written as 𝔪 = 𝔪0𝔪∞ by separating the finite from the (real) infinite
primes.
We say a ≡ b (mod 𝔪) if
If 𝔭 is a finite prime, then a ≡ b (mod 𝔭ν(𝔭)) means exactly what you think it
should mean: a − b ∈𝔭ν(𝔭).
If 𝔭 is a real infinite prime σ : K → ℝ, then a ≡ b (mod 𝔭) means that σ(a∕b) > 0.
Of course, a ≡ b (mod 𝔪) means a ≡ b modulo each prime power in 𝔪. With this, we can define
a generalization of the class group:
Definition 56.2.1.Let 𝔪 be a modulus of a number field K.
Let IK(𝔪) denote the set of all fractional ideals of K which are relatively prime
to 𝔪, which is an abelian group.
Let PK(𝔪) be the subgroup of IK(𝔪) generated by
This is sometimes called a “ray” of principal ideals.
Finally define the ray class group of 𝔪 to be CK(𝔪) = IK(𝔪)∕PK(𝔪).
This group is known to always be finite. Note the usual class group is CK(1).
One last definition that we’ll use right after Artin reciprocity:
Definition 56.2.2.A congruence subgroup of 𝔪 is a subgroup H with
Thus CK(𝔪) is a group which contains a lattice of various quotients IK(𝔪)∕H, where H is a
congruence subgroup.
This definition takes a while to get used to, so here are examples.
Example 56.2.3 (Ray class groups in ℚ, finite modulus) Consider K = ℚ with infinite prime
∞. Then
If we take 𝔪 = 1 then Iℚ(1) is all fractional ideals, and Pℚ(1) is all principal
fractional ideals. Their quotient is the usual class group of ℚ, which is trivial.
Now take 𝔪 = 8. Thus Iℚ(8). Moreover
You might at first glance think that the quotient is thus (ℤ∕8ℤ)×. But the issue is
that we are dealing with ideals: specifically, we have
because −7 ≡ 1 (mod 8). So actually, we get
More generally,
Example 56.2.4 (Ray class groups in ℚ, infinite moduli) Consider K = ℚ with infinite prime
∞ again.
Now take 𝔪 = ∞. As before Iℚ(∞) = ℚ×. Now, by definition we have
At first glance you might think this was ℚ>0, but the same behavior with ideals
shows in fact Pℚ(∞) = ℚ×. So in this case, Pℚ(∞) still has all principal fractional
ideals. Therefore, Cℚ(∞) is still trivial.
Finally, let 𝔪 = 8∞. As before Iℚ(8∞). Now in
this case:
This time, we really do have −7ℤPℚ(8∞): we have 7≢1 (mod 8) and also
−8 < 0. So neither of the generators of 7ℤ are in Pℚ(8∞). Thus we finally obtain
with the bijection Cℚ(8∞) → (ℤ∕8ℤ)× given by aℤ|a| (mod 8).
More generally,
56.3Infinite primes in extensions
I want to emphasize that everything above is intrinsic to a particular number field K. After
this point we are going to consider extensions L∕K but it is important to keep in mind the
distinction that the concept of modulus and ray class group are objects defined solely from K
rather than the above L.
Now take a Galois extension L∕K of degree m. We already know prime ideals 𝔭 of K
break into a product of prime ideals 𝔓 of K in L in a nice way, so we want to do
the same thing with infinite primes. This is straightforward: each of the n infinite
primes σ : K → ℂ lifts to m infinite primes τ : L → ℂ, by which I mean the diagram
commutes. Hence like before, each infinite prime σ of K has m infinite primes τ of L which lie
above it.
For a real prime σ of K, any of the resulting τ above it are complex, we say that the
prime σ ramifies in the extension L∕K. Otherwise it is unramified in L∕K. An
infinite prime of K is always unramified in L∕K. In this way, we can talk about an
unramified Galois extension L∕K: it is one where all primes (finite or infinite) are
unramified.
Example 56.3.1 (Ramification of ∞) Let ∞ be the real infinite prime of ℚ.
∞ is ramified in ℚ()∕ℚ.
∞ is unramified in ℚ()∕ℚ.
Note also that if K is totally complex then any extension L∕K is unramified.
56.4Frobenius element and Artin symbol
Recall the key result:
Theorem 56.4.1 (Frobenius element) Let L∕K be a Galois extension. If 𝔭 is a prime
unramified in K, and 𝔓 a prime above it in L. Then there is a unique element of
Gal(L∕K), denoted Frob𝔓, obeying
Example 56.4.2 (Example of Frobenius elements) Let L = ℚ(i), K = ℚ. We have
Gal(L∕K)ℤ∕2ℤ.
If p is an odd prime with 𝔓 above it, then Frob𝔓 is the unique element such that
in ℤ[i]. In particular,
From this we see that Frob𝔓 is the identity when p ≡ 1 (mod 4) and Frob𝔓 is complex
conjugation when p ≡ 3 (mod 4).
Example 56.4.3 (Cyclotomic Frobenius element) Generalizing previous example, let
L = ℚ(ζ) and K = ℚ, with ζ an mth root of unity. It’s well-known that L∕K is
unramified outside ∞ and prime factors of m. Moreover, the Galois group Gal(L∕K) is
(ℤ∕mℤ)×: the Galois group consists of elements of the form
and Gal(L∕K) = .
Then it follows just like before that if p ∤ n is prime and 𝔓 is above p
An important property of the Frobenius element is its order is related to the decomposition
of 𝔭 in the higher field L in the nicest way possible:
Lemma 56.4.4 (Order of the Frobenius element) The Frobenius
element Frob𝔓∈ Gal(L∕K) of an extension L∕K has order equal to the inertial degree
of 𝔓, that is,
In particular, Frob𝔓 = id if and only if 𝔭 splits completely in L∕K.
Proof.We want to understand the order of the map T : xxN𝔭 on the field 𝒪K∕𝔓.
But the latter is isomorphic to the splitting field of XN𝔓−X in 𝔽p, by Galois theory of
finite fields. Hence the order is log N𝔭(N𝔓) = f(𝔓∣𝔭). □
Exercise 56.4.5.Deduce from this that the rational primes which split completely in ℚ(ζ) areexactly those with p ≡ 1 (modm). Here ζ is an mth root of unity.
The Galois group acts transitively among the set of 𝔓 above a given 𝔭, so that we
have
Thus Frob𝔓 is determined by its underlying 𝔭 up to conjugation.
In class field theory, we are interested in abelian extensions, i.e. those for which
Gal(L∕K) is Galois. Here the theory becomes extra nice: the conjugacy classes have size one.
Definition 56.4.6.Assume L∕K is an abelian extension. Then for a given unramified
prime 𝔭 in K, the element Frob𝔓 doesn’t depend on the choice of 𝔓. We denote the
resulting Frob𝔓 by the Artin symbol,
The definition of the Artin symbol is written deliberately to look like the Legendre symbol. To
see why:
Example 56.4.7 (Legendre symbol subsumed by Artin symbol) Suppose we want to
understand (2∕p) ≡ 2 where p > 2 is prime. Consider the element
It is uniquely determined by where it sends a. But in fact we have
where is the usual Legendre symbol, and 𝔓 is above p in ℚ(). Thus the Artin
symbol generalizes the quadratic Legendre symbol.
Example 56.4.8 (Cubic Legendre symbol subsumed by Artin symbol) Similarly, it also
generalizes the cubic Legendre symbol. To see this, assume 𝜃 is primary in K =
ℚ() = ℚ(ω) (thus 𝒪K = ℤ[ω] is Eisenstein integers). Then for example
where 𝔓 is above p in K().
56.5Artin reciprocity
Now, we further capitalize on the fact that Gal(L∕K) is abelian. For brevity, in what follows
let Ram(L∕K) denote the primes of K (either finite or infinite) which ramify in
L.
Definition 56.5.1.Let L∕K be an abelian extension and let 𝔪 be divisible by every
prime in Ram(L∕K). Then since L∕K is abelian we can extend the Artin symbol
multiplicatively to a map
This is called the Artin map, and it is surjective (for example by Chebotarev
Density). Thus we denote its kernel by
In particular we have Gal(L∕K)IK(𝔪)∕H(L∕K,𝔪).
We can now present the long-awaited Artin reciprocity theorem.
Theorem 56.5.2 (Artin reciprocity) Let L∕K be an abelian extension. Then there is a
modulus 𝔣 = 𝔣(L∕K), divisible by exactly the primes of Ram(L∕K), such that: for any
modulus 𝔪 divisible by all primes of Ram(L∕K), we have
We call 𝔣 the conductor of L∕K.
So the conductor 𝔣 plays a similar role to the discriminant (divisible by exactly the primes
which ramify), and when 𝔪 is divisible by the conductor, H(L∕K,𝔪) is a congruencesubgroup.
Here’s the reason this is called a “reciprocity” theorem. Recalling that CK(𝔣) = IK(𝔣)∕PK(𝔣),
the above theorem tells us we get a sequence of maps
Consequently:
For primes 𝔭 ∈ IK(𝔣), depends only on “𝔭 (mod 𝔣)”.
Let’s see how this result relates to quadratic reciprocity.
Example 56.5.3 (Artin reciprocity implies quadratic reciprocity) The big miracle of
quadratic reciprocity states that: for a fixed (squarefree) a, the Legendre symbol
should only depend the residue of p modulo something. Let’s see why Artin reciprocity
tells us this a priori.
Let L = ℚ(), K = ℚ. Then we’ve already seen that the Artin symbol
is the correct generalization of the Legendre symbol. Thus, Artin reciprocity tells us that
there is a conductor 𝔣 = 𝔣(ℚ()∕ℚ) such that depends only on the residue
of p modulo 𝔣, which is what we wanted.
Here is an example along the same lines.
Example 56.5.4 (Cyclotomic field) Let ζ be a primitive mth root of unity. For primes
p, we know that Frobp∈ Gal(ℚ(ζ)∕ℚ) is “exactly” p (mod m). Let’s translate this idea
into the notation of Artin reciprocity.
We are going to prove
This is the generic example of achieving the lower bound in Artin reciprocity. It also
implies that 𝔣(ℚ(ζ)∕ℚ)∣m∞.
It’s well-known ℚ(ζ)∕ℚ is unramified outside finite primes dividing m, so that the Artin
symbol is defined on IK(𝔪). Now the Artin map is given by
So we see that the kernel of this map is trivial, i.e. it is given by the identity of the Galois
group, corresponding to 1 (mod m). On the other hand, we’ve also computed Pℚ(m∞)
already, so we have the desired equality.
In fact, we also have the following “existence theorem”: every congruence subgroup appears
uniquely once we fix 𝔪.
Theorem 56.5.5 (Takagi existence theorem) Fix K and let 𝔪 be a modulus. Consider any
congruence subgroup H, i.e.
Then H = H(L∕K,𝔪) for a unique abelian extension L∕K.
Finally, such subgroups reverse inclusion in the best way possible:
Lemma 56.5.6 (Inclusion-reversing congruence subgroups) Fix a modulus 𝔪. Let L∕K and
M∕K be abelian extensions and suppose 𝔪 is divisible by the conductors of L∕K and
M∕K. Then
Here by L ⊆ M we mean that L is isomorphic to some subfield of M.
Sketch of proof. Let us first prove the equivalence with 𝔪 fixed. In one direction, assume
L ⊆ M; one can check from the definitions that the diagram
commutes, because it suffices to verify this for prime powers, which is just saying that
Frobenius elements behave well with respect to restriction. Then the inclusion of
kernels follows directly. The reverse direction is essentially the Takagi existence
theorem. □
Note that we can always take 𝔪 to be the product of conductors here.
To finish, here is a quote from Emil Artin on his reciprocity law:
I will tell you a story about the Reciprocity Law. After my thesis, I had the
idea to define L-series for non-abelian extensions. But for them to agree with
the L-series for abelian extensions, a certain isomorphism had to be true.
I could show it implied all the standard reciprocity laws. So I called it the
General Reciprocity Law and tried to prove it but couldn’t, even after many
tries. Then I showed it to the other number theorists, but they all laughed
at it, and I remember Hasse in particular telling me it couldn’t possibly be
true.
Still, I kept at it, but nothing I tried worked. Not a week went by — for threeyears! — that I did not try to prove the Reciprocity Law. It was discouraging,
and meanwhile I turned to other things. Then one afternoon I had nothing
special to do, so I said, ‘Well, I try to prove the Reciprocity Law again.’ So
I went out and sat down in the garden. You see, from the very beginning I
had the idea to use the cyclotomic fields, but they never worked, and now I
suddenly saw that all this time I had been using them in the wrong way —
and in half an hour I had it.
56.6A few harder problems to think about
Problem 56A† (Kronecker-Weber theorem).Let L be an abelian extension of ℚ. Then L is
contained in a cyclic extension ℚ(ζ) where ζ is an mth root of unity (for some m).
Problem 56B† (Hilbert class field).Let K be any number field. Then there exists a unique
abelian extension E∕K which is unramified at all primes (finite or infinite) and such that
E∕K is the maximal such extension by inclusion.
Gal(E∕K) is isomorphic to the class group of E.
A prime 𝔭 of K splits completely in E if and only if it is a principal ideal of 𝒪K.
We call E the Hilbert class field of K.
Part XV Algebraic Topology I: Homotopy
57Some topological constructions
In this short chapter we briefly describe some common spaces and constructions in topology
that we haven’t yet discussed.
57.1Spheres
Recall that
is the surface of an n-sphere while
is the corresponding closed ball (So for example, D2 is a disk in a plane while S1 is the unit
circle.)
Exercise 57.1.1.Show that the open ball Dn∖ Sn−1is homeomorphic to ℝn.
In particular, S0 consists of two points, while D1 can be thought of as the interval
[−1,1].
57.2Quotient topology
Prototypical example for this section:Dn∕Sn−1 = Sn, or the torus.
Given a space X, we can identify some of the points together by any equivalence relation ∼;
for an x ∈ X we denote its equivalence class by [x]. Geometrically, this is the space achieved
by welding together points equivalent under ∼.
Formally,
Definition 57.2.1.Let X be a topological space, and ∼ an equivalence relation on the points
of X. Then X∕∼ is the space whose
Points are equivalence classes of X, and
U ⊆ X∕∼ is open if and only if is open in X.
As far as I can tell, this definition is mostly useless for intuition, so here are some
examples.
Example 57.2.2 (Interval modulo endpoints) Suppose we take D1 = [−1,1] and
quotient by the equivalence relation which identifies the endpoints −1 and 1. (Formally,
x ∼ y⇔ (x = y) or {x,y} = {−1,1}.) In that case, we simply recover S1:
Observe that a small open neighborhood around −1 ∼ 1 in the quotient space corresponds to
two half-intervals at −1 and 1 in the original space D1. This should convince you the
definition we gave is the right one.
Example 57.2.3 (More quotient spaces) Convince yourself that:
Generalizing the previous example, Dn modulo its boundary Sn−1 is Sn.
Given a square ABCD, suppose we identify segments AB and DC together. Then
we get a cylinder. (Think elementary school, when you would tape up pieces of
paper together to get cylinders.)
In the previous example, if we also identify BC and DA together, then we get a
torus. (Imagine taking our cylinder and putting the two circles at the end together.)
Let X = ℝ, and let x ∼ y if y − x ∈ ℤ. Then X∕∼ is S1 as well.
One special case that we did above:
Definition 57.2.4.Let A ⊆ X. Consider the equivalence relation which identifies all
the points of A with each other while leaving all remaining points inequivalent. (In other
words, x ∼ y if x = y or x,y ∈ A.) Then the resulting quotient space is denoted X∕A.
So in this notation,
Abuse of Notation 57.2.5.Note that I’m deliberately being sloppy, and saying
“Dn∕Sn−1 = Sn” or “Dn∕Sn−1is Sn”, when I really ought to say “Dn∕Sn−1 is
homeomorphic to Sn”. This is a general theme in mathematics: objects which are
homoeomorphic/isomorphic/etc. are generally not carefully distinguished from each
other.
57.3Product topology
Prototypical example for this section:ℝ × ℝ is ℝ2, S1× S1is the torus.
Definition 57.3.1.Given topological spaces X and Y , the product topology on X ×Y is
the space whose
Points are pairs (x,y) with x ∈ X, y ∈ Y , and
Topology is given as follows: the basis of the topology for X × Y is U × V , for
U ⊆ X open and V ⊆ Y open.
Remark 57.3.2 — It is not hard to show that, in fact, one need only consider basis
elements for U and V . That is to say,
is also a basis for X × Y .
We really do need to fiddle with the basis: in ℝ × ℝ, an open unit disk better be open,
despite not being of the form U × V .
This does exactly what you think it would.
Example 57.3.3 (The unit square) Let X = [0,1] and consider X × X. We of course expect
this to be the unit square. Pictured below is an open set of X × X in the basis.
Exercise 57.3.4.Convince yourself this basis gives the same topology as the product metric onX × X. So this is the “right” definition.
Example 57.3.5 (More product spaces)
(a)
ℝ × ℝ is the Euclidean plane.
(b)
S1× [0,1] is a cylinder.
(c)
S1× S1 is a torus! (Why?)
57.4Disjoint union and wedge sum
Prototypical example for this section:S1∨ S1is the figure eight.
The disjoint union of two spaces is geometrically exactly what it sounds like: you just
imagine the two spaces side by side. For completeness, here is the formal definition.
Definition 57.4.1.Let X and Y be two topological spaces. The disjoint union, denoted
X ∐ Y , is defined by
The points are the disjoint union X ∐ Y , and
A subset U ⊆ X ∐ Y is open if and only if U ∩ X and U ∩ Y are open.
Exercise 57.4.2.Show that the disjoint union of two nonempty spaces is disconnected.
More interesting is the wedge sum, where two topological spaces X and Y are fused together
only at a single base point.
Definition 57.4.3.Let X and Y be topological spaces, and x0∈ X and y0∈ Y be
points. We define the equivalence relation ∼ by declaring x0∼ y0 only. Then the wedgesum of two spaces is defined as
Example 57.4.4 (S1∨ S1 is a figure eight) Let X = S1 and Y = S1, and let x0∈ X and
y0∈ Y be any points. Then X ∨ Y is a “figure eight”: it is two circles fused together at one
point.
Abuse of Notation 57.4.5.We often don’t mention x0 and y0 when they are
understood (or irrelevant). For example, from now on we will just write S1∨ S1 for a
figure eight.
Remark 57.4.6 — Annoyingly, in LaTeX \wedge gives ∧ instead of ∨ (which is \vee).
So this really should be called the “vee product”, but too late.
57.5CW complexes
Using this construction, we can start building some spaces. One common way to do so
is using a so-called CW complex. Intuitively, a CW complex is built as follows:
Start with a set of points X0.
Define X1 by taking some line segments (copies of D1) and fusing the endpoints
(copies of S0) onto X0.
Define X2 by taking copies of D2 (a disk) and welding its boundary (a copy of S1)
onto X1.
Repeat inductively up until a finite stage n; we say X is n-dimensional.
The resulting space X is the CW-complex. The set Xk is called the k-skeleton of X. Each Dk is
called a k-cell; it is customary to denote it by eαk where α is some index. We say that X is
finite if only finitely many cells were used.
Abuse of Notation 57.5.1.Technically, most sources (like [?]) allow one to construct
infinite-dimensional CW complexes. We will not encounter any such spaces in the
Napkin.
Example 57.5.2 (D2 with 2 + 2 + 1 and 1 + 1 + 1 cells)
(a)
First, we start with X0 having two points ea0 and eb0. Then, we join them with two
1-cells D1 (green), call them ec1 and ed1. The endpoints of each 1-cell (the copy of
S0) get identified with distinct points of X0; hence X1S1. Finally, we take a single
2-cell e2 (yellow) and weld it in, with its boundary fitting into the copy of S1 that
we just drew. This gives the figure on the left.
(b)
In fact, one can do this using just 1 + 1 + 1 = 3 cells. Start with X0 having a single
point e0. Then, use a single 1-cell e1, fusing its two endpoints into the single point
of X0. Then, one can fit in a copy of S1 as before, giving D2 as on the right.
Example 57.5.3 (Sn as a CW complex)
(a)
One can obtain Sn (for n ≥ 1) with just two cells. Namely, take a single point e0 for
X0, and to obtain Sn take Dn and weld its entire boundary into e0.
We already saw this example in the beginning with n = 2, when we saw that the
sphere S2 was the result when we fuse the boundary of a disk D2 together.
(b)
Alternatively, one can do a “hemisphere” construction, by constructing Sn
inductively using two cells in each dimension. So S0 consists of two points, then S1
is obtained by joining these two points by two segments (1-cells), and S2 is obtained
by gluing two hemispheres (each a 2-cell) with S1 as its equator.
Definition 57.5.4.Formally, for each k-cell eαk we want to add to Xk, we take its
boundary Sαk−1 and weld it onto Xk−1 via an attaching mapSαk−1→ Xk−1. Then
where ∼ identifies each boundary point of eαk with its image in Xk−1.
57.6The torus, Klein bottle, ℝℙn, ℂℙn
We now present four of the most import examples of CW complexes.
57.6.iThe torus
The torus can be formed by taking a square and identifying the opposite edges in the same
direction: if you walk off the right edge, you re-appear at the corresponding point in on the left
edge. (Think Asteroids from Atari!)
Thus the torus is (ℝ∕ℤ)2S1× S1.
Note that all four corners get identified together to a single point. One can realize the torus
in 3-space by treating the square as a sheet of paper, taping together the left and right (red)
edges to form a cylinder, then bending the cylinder and fusing the top and bottom (blue)
edges to form the torus.
Image from [?]
The torus can be realized as a CW complex with
A 0-skeleton consisting of a single point,
A 1-skeleton consisting of two 1-cells ea1, eb1, and
A 2-skeleton with a single 2-cell e2, whose circumference is divided into four parts, and
welded onto the 1-skeleton “via aba−1b−1”. This means: wrap a quarter of the
circumference around ea1, then another quarter around eb1, then the third quarter
around ea1 but in the opposite direction, and the fourth quarter around eb1 again in the
opposite direction as before.
We say that aba−1b−1 is the attaching word; this shorthand will be convenient later
on.
57.6.iiThe Klein bottle
The Klein bottle is defined similarly to the torus, except one pair of edges is identified in the
opposite manner, as shown.
Unlike the torus one cannot realize this in 3-space without self-intersecting. One can tape
together the red edges as before to get a cylinder, but to then fuse the resulting blue
circles in opposite directions is not possible in 3D. Nevertheless, we often draw a
picture in 3-dimensional space in which we tacitly allow the cylinder to intersect
itself.
Image from [?,?]
Like the torus, the Klein bottle is realized as a CW complex with
One 0-cell,
Two 1-cells ea1 and eb1, and
A single 2-cell attached this time via the word abab−1.
57.6.iiiReal projective space
Let’s start with n = 2. The space ℝℙ2 is obtained if we reverse both directions of the square
from before, as shown.
However, once we do this the fact that the original polygon is a square is kind of irrelevant;
we can combine a red and blue edge to get the single purple edge. Equivalently, one
can think of this as a circle with half its circumference identified with the other
half:
The resulting space should be familiar to those of you who do projective (Euclidean)
geometry. Indeed, there are several possible geometric interpretations:
One can think of ℝℙ2 as the set of lines through the origin in ℝ3, with each line
being a point in ℝℙ2.
Of course, we can identify each line with a point on the unit sphere S2, except
for the property that two antipodal points actually correspond to the same line,
so that ℝℙ2 can be almost thought of as “half a sphere”. Flattening it gives the
picture above.
Imagine ℝ2, except augmented with “points at infinity”. This means that we add
some points “infinitely far away”, one for each possible direction of a line. Thus in
ℝℙ2, any two lines indeed intersect (at a Euclidean point if they are not parallel,
and at a point at infinity if they do).
This gives an interpretation of ℝℙ2, where the boundary represents the line atinfinity through all of the points at infinity. Here we have used the fact that ℝ2
and interior of D2 are homeomorphic.
Exercise 57.6.1.Observe that these formulations are equivalent by considering the plane z = 1in ℝ3, and intersecting each line in the first formulation with this plane.
We can also express ℝℙ2 using coordinates: it is the set of triples (x : y : z) of real numbers
not all zero up to scaling, meaning that
for any λ≠0. Using the “lines through the origin in ℝ3” interpretation makes it clear why
this coordinate system gives the right space. The points at infinity are those with z = 0, and
any point with z≠0 gives a Cartesian point since
hence we can think of it as the Cartesian point (,).
In this way we can actually define real-projective n-space, ℝℙn for any n, as
either
(i)
The set of lines through the origin in ℝn+1,
(ii)
Using n + 1 coordinates as above, or
(iii)
As ℝn augmented with points at infinity, which themselves form a copy of ℝℙn−1.
As a possibly helpful example, we give all three pictures of ℝℙ1.
Example 57.6.2 (Real projective 1-Space) ℝℙ1 can be thought of as S1 modulo the relation
the antipodal points are identified. Projecting onto a tangent line, we see that we get a copy of
ℝ plus a single point at infinity, corresponding to the parallel line (drawn in cyan below).
Thus, the points of ℝℙ1 have two forms:
(x : 1), which we think of as x ∈ ℝ (in dark red above), and
(1 : 0), which we think of as 1∕0 = ∞, corresponding to the cyan line above.
So, we can literally write
Note that ℝℙ1 is also the boundary of ℝℙ2. In fact, note also that topologically we
have
since it is the “real line with endpoints fused together”.
Since ℝℙn is just “ℝn (or Dn) with ℝℙn−1 as its boundary”, we can construct ℝℙn
as a CW complex inductively. Note that ℝℙn thus consists of one cell in eachdimension.
Example 57.6.3 (ℝℙn as a cell complex)
(a)
ℝℙ0 is a single point.
(b)
ℝℙ1S1 is a circle, which as a CW complex is a 0-cell plus a 1-cell.
(c)
ℝℙ2 can be formed by taking a 2-cell and wrapping its perimeter twice around a
copy of ℝℙ1.
57.6.ivComplex projective space
The complex projective spaceℂℙn is defined like ℝℙn with coordinates, i.e.
under scaling; this time zi are complex. As before, ℂℙn can be thought of as ℂn augmented
with some points at infinity (corresponding to ℂℙn−1).
Example 57.6.4 (Complex projective space)
(a)
ℂℙ0 is a single point.
(b)
ℂℙ1 is ℂ plus a single point at infinity (“complex infinity” if you will). That means
as before we can think of ℂℙ1 as
So, imagine taking the complex plane and then adding a single point to encompass
the entire boundary. The result is just sphere S2.
Here is a picture of ℂℙ1 with its coordinate system, the Riemann sphere.
Remark 57.6.5 (For Euclidean geometers) — You may recognize that while ℝℙ2 is the
setting for projective geometry, inversion about a circle is done in ℂℙ1 instead. When
one does an inversion sending generalized circles to generalized circles, there is only one
point at infinity: this is why we work in ℂℙn.
Like ℝℙn, ℂℙn is a CW complex, built inductively by taking ℂn and welding its boundary
onto ℂℙn−1 The difference is that as topological spaces,
Thus, we attach the cells D0, D2, D4 and so on inductively to construct ℂℙn. Thus we see
that
ℂℙnconsists of one cell in each even dimension.
57.7A few harder problems to think about
Problem 57A.Show that a space X is Hausdorff if and only if the diagonal {(x,x)∣x ∈X} is closed in the product space X × X.
Problem 57B.Realize the following spaces as CW complexes:
(a)
Möbius strip.
(b)
ℝ.
(c)
ℝn.
Problem 57C†.Show that a finite CW complex is compact.
58Fundamental groups
Topologists can’t tell the difference between a coffee cup and a doughnut. So how do you tell
anything apart?
This is a very hard question to answer, but one way we can try to answer it is to find some
invariants of the space. To draw on the group analogy, two groups are clearly not isomorphic
if, say, they have different orders, or if one is simple and the other isn’t, etc. We’d like to find
some similar properties for topological spaces so that we can actually tell them
apart.
Two such invariants for a space X are
Defining homology groups H1(X), H2(X), …
Defining homotopy groups π1(X), π2(X), …
Homology groups are hard to define, but in general easier to compute. Homotopy groups are
easier to define but harder to compute.
This chapter is about the fundamental group π1.
58.1Fusing paths together
Recall that a path in a space X is a function [0,1] → X. Suppose
we have paths γ1 and γ2 such that γ1(1) = γ2(0). We’d like to
fuse1
them together to get a path γ1∗ γ2. Easy, right?
We unfortunately do have to hack the definition a tiny bit. In an ideal world,
we’d have a path γ1 : [0,1] → X and γ2 : [1,2] → X and we could just merge them
together to get γ1∗ γ2 : [0,2] → X. But the “2” is wrong here. The solution is that we
allocate [0,] for the first path and [,1] for the second path; we run “twice as
fast”.
Definition 58.1.1.Given two paths γ1,γ2 : [0,1] → X such that γ1(1) = γ2(0), we
define a path γ1∗ γ2 : [0,1] → X by
This hack unfortunately reveals a second shortcoming: this “product” is not associative. If
we take (γ1∗ γ2) ∗ γ3 for some suitable paths, then [0,], [,] and [,1] are the times
allocated for γ1, γ2, γ3.
Question 58.1.2.What are the times allocated for γ1∗ (γ2∗ γ3)?
But I hope you’ll agree that even though this operation isn’t associative, the reason it fails
to be associative is kind of stupid. It’s just a matter of how fast we run in certain
parts.
So as long as we’re fusing paths together, we probably don’t want to think of [0,1] itself too
seriously. And so we only consider everything up to (path) homotopy equivalence. (Recall
that two paths α and β are homotopic if there’s a path homotopy F : [0,1]2→ X
between them, which is a continuous deformation from α to β.) It is definitely true
that
It is also true that if α1≃ α2 and β1≃ β2 then α1∗ β1≃ α2∗ β2.
Naturally, homotopy is an equivalence relation, so paths γ lives in some “homotopy
type”, the equivalence classes under ≃. We’ll denote this [γ]. Then it makes sense to
talk about [α] ∗ [β]. Thus, we can think of ∗ as an operation on homotopyclasses.
58.2Fundamental groups
Prototypical example for this section:π1(ℝ2) is trivial and π1(S1)ℤ.
At this point I’m a little annoyed at keeping track of endpoints, so now I’m going to
specialize to a certain type of path.
Definition 58.2.1.A loop is a path with γ(0) = γ(1).
Hence if we restrict our attention to paths starting at a single point x0,
then we can stop caring about endpoints and start-points, since everything
starts and stops at x0. We even have a very canonical loop: the “do-nothing”
loop2
given by standing at x0 the whole time.
Definition 58.2.2.Denote the trivial “do-nothing loop” by 1. A loop γ is
nulhomotopic if it is homotopic to 1; i.e. γ ≃ 1.
For homotopy of loops, you might visualize “reeling in” the loop, contracting it to a single
point.
Example 58.2.3 (Loops in S2 are nulhomotopic) As the following picture should convince
you, every loop in the simply connected space S2 is nulhomotopic.
(Starting with the purple loop, we contract to the red-brown point.)
Hence to show that spaces are simply connected it suffices to understand the loops of that
space. We are now ready to provide:
Definition 58.2.4.The fundamental group of X with basepoint x0, denoted
π1(X,x0), is the set of homotopy classes
equipped with ∗ as a group operation.
It might come as a surprise that this has a group structure. For example, what is the
inverse? Let’s define it now.
Definition 58.2.5.Given a path α : [0,1] → X we can define a path α
In effect, this “runs α backwards”. Note that α starts at the endpoint of α and ends
at the starting point of α.
Exercise 58.2.6.Show that for any path α, α ∗αis homotopic to the “do-nothing” loop atα(0). (Draw a picture.)
Let’s check it.
Proof that this is a group structure. Clearly ∗ takes two loops at x0 and spits out a loop at
x0. We also already took the time to show that ∗ is associative. So we only have to check that
(i) there’s an identity, and (ii) there’s an inverse.
We claim that the identity is the “do-nothing” loop 1 we described above. The
reader can check that for any γ,
For a loop γ, recall again we define its “backwards” loop γ by
Then we have γ ∗γ =γ∗ γ = 1.
Hence π1(X,x0) is actually a group. □
Before going any further I had better give some examples.
Example 58.2.7 (Examples of fundamental groups) Note that proving the following results is
not at all trivial. For now, just try to see intuitively why the claimed answer “should” be
correct.
(a)
The fundamental group of ℂ is the trivial group: in the plane, every loop is
nulhomotopic. (Proof: imagine it’s a piece of rope and reel it in.)
(b)
On the other hand, the fundamental group of ℂ−{0} (meteor example from earlier)
with any base point is actually ℤ! We won’t be able to prove this for a while, but
essentially a loop is determined by the number of times that it winds around the
origin – these are so-called winding numbers. Think about it!
(c)
Similarly, we will soon show that the fundamental group of S1 (the boundary of the
unit circle) is ℤ.
Officially, I also have to tell you what the base point is, but by symmetry in these examples, it
doesn’t matter.
Here is the picture for ℂ ∖{0}, with the hole exaggerated as the meteor from ?? .
Question 58.2.8.Convince yourself that the fundamental group of S1is ℤ, and understand whywe call these “winding numbers”. (This will be the most important example of a fundamentalgroup in later chapters, so it’s crucial you figure it out now.)
Example 58.2.9 (The figure eight) Consider a figure eight S1∨ S1, and let x0 be the
center. Then
is the free group generated on two letters. The idea is that one loop of the eight is a, and the
other loop is b, so we expect π1 to be generated by this loop a and b (and its inverses a and b).
These loops don’t talk to each other.
Recall that in graph theory, we usually assume our graphs are connected, since
otherwise we can just consider every connected component separately. Likewise, we
generally want to restrict our attention to path-connected spaces, since if a space isn’t
path-connected then it can be broken into a bunch of “path-connected components”. (Can you
guess how to define this?) Indeed, you could imagine a space X that consists of
the objects on my desk (but not the desk itself): π1 of my phone has nothing to
do with π1 of my mug. They are just totally disconnected, both figuratively and
literally.
But on the other hand we claim that in a path-connected space, the groups are very
related!
Theorem 58.2.10 (Fundamental groups don’t depend on basepoint) Let X be a
path-connected space. Then for any x1∈ X and x2∈ X, we have
Before you read the proof, see if you can guess the isomorphism based just on the picture
below.
Proof.Let α be any path from x1 to x2 (possible by path-connectedness), and let α be
its reverse. Then we can construct a map
In other words, given a loop γ at x1, we can start at x2, follow α to x1, run γ, then
run along α home to x2. Hence this is a map which builds a loop of π1(X,x2) from every
loop at π1(X,x1). It is a homomorphism of the groups just because
as α ∗α is nulhomotopic.
Similarly, there is a homomorphism
As these maps are mutual inverses, it follows they must be isomorphisms. End of
story. □
This is a bigger reason why we usually only care about path-connected spaces.
Abuse of Notation 58.2.11.For a path-connected space X we will often abbreviate
π1(X,x0) to just π1(X), since it doesn’t matter which x0∈ X we pick.
Finally, recall that we originally defined “simply connected” as saying that any two paths
with matching endpoints were homotopic. It’s possible to weaken this condition and then
rephrase it using fundamental groups.
Exercise 58.2.12.Let X be a path-connected space. Prove that X is simply connectedifand only if π1(X) is the trivial group. (One direction is easy; the other is a little trickier.)
This is the “usual” definition of simply connected.
58.3Fundamental groups are functorial
One quick shorthand I will introduce to clean up the discussion:
Definition 58.3.1.By f : (X,x0) → (Y,y0), we will mean that f : X → Y is a
continuous function of spaces which also sends the point x0 to y0.
Let X and Y be topological spaces and f : (X,x0) → (Y,y0). We now want to relate the
fundamental groups of X and Y .
Recall that a loop γ in (X,x0) is a map γ : [0,1] → X with γ(0) = γ(1) = x0. Then if we
consider the composition
then we get straight-away a loop in Y at y0! Let’s call this loop f♯γ.
Lemma 58.3.2 (f♯is homotopy invariant) If γ1≃ γ2 are path-homotopic, then in fact
Proof.Just take the homotopy h taking γ1 to γ2 and consider f ∘ h. □
It’s worth noting at this point that if X and Y are homeomorphic, then their
fundamental groups are all isomorphic. Indeed, let f : X → Y and g : Y → X be mutually
inverse continuous maps. Then one can check that f♯ : π1(X,x0) → π1(Y,y0) and
g♯ : π1(Y,y0) → π1(X,x0) are inverse maps between the groups (assuming f(x0) = y0 and
g(y0) = x0).
58.4Higher homotopy groups
Why the notation π1 for the fundamental group? And what are π2, …? The answer lies in the
following rephrasing:
Question 58.4.1.Convince yourself that a loop is the same thing as a continuous functionS1→ X.
It turns out we can define homotopy for things other than paths. Two functions
f,g : Y → X are homotopic if there exists a continuous function Y × [0,1] → X which
continuously deforms f to g. So everything we did above was just the special case
Y = S1.
For general n, the group πn(X) is defined as the homotopy classes of the maps Sn→ X.
The group operation is a little harder to specify. You have to show that Sn is homeomorphic
to [0,1]n with some endpoints fused together; for example S1 is [0,1] with 0 fused to 1. Once
you have these cubes, you can merge them together on a face. (Again, I’m being terribly
imprecise, deliberately.)
For n≠1, πn behaves somewhat differently than π1. (You might not be surprised, as Sn is
simply connected for all n ≥ 2 but not when n = 1.) In particular, it turns out that πn(X) is
an abelian group for all n ≥ 2.
Let’s see some examples.
Example 58.4.2 (πn(Sn)ℤ) As we saw, π1(S1)ℤ; given the base circle S1, we can
wrap a second circle around it as many times as we want. In general, it’s true that
πn(Sn)ℤ.
Example 58.4.3 (πn(Sm){1} when n < m) We saw that π1(S2){1}, because a
circle in S2 can just be reeled in to a point. It turns out that similarly, any smaller
n-dimensional sphere can be reeled in on the surface of a bigger m-dimensional sphere.
So in general, πn(Sm) is trivial for n < m.
However, beyond these observations, the groups behave quite weirdly. Here is a table of
πn(Sm) for 1 ≤ m ≤ 8 and 2 ≤ n ≤ 10, so you can see what I’m talking about. (Taken from
Wikipedia.)
Actually, it turns out that if you can compute πn(Sm) for every m and n, then you can
essentially compute any homotopy classes. Thus, computing πn(Sm) is sort of a lost cause in
general, and the mixture of chaos and pattern in the above table is a testament to
this.
58.5Homotopy equivalent spaces
Prototypical example for this section:A disk is homotopy equivalent to a point, an annulus ishomotopy equivalent to S1.
Up to now I’ve abused notation and referred to “path homotopy” as just “homotopy” for
two paths. I will unfortunately continue to do so (and so any time I say two paths are
homotopic, you should assume I mean “path-homotopic”). But let me tell you what the
general definition of homotopy is first.
Definition 58.5.1.Let f,g : X → Y be continuous functions. A homotopy is a
continuous function F : X × [0,1] → Y , which we’ll write Fs(x) for s ∈ [0,1], x ∈ X,
such that
If such a function exists, then f and g are homotopic.
Intuitively this is once again “deforming f to g”. You might notice this is almost exactly the
same definition as path-homotopy, except that f and g are any functions instead of paths,
and hence there’s no restriction on keeping some “endpoints” fixed through the
deformation.
This homotopy can be quite dramatic:
Example 58.5.2 The zero function z0 and the identity function zz are homotopic
as functions ℂ → ℂ. The necessary deformation is
I bring this up because I want to define:
Definition 58.5.3.Let X and Y be continuous spaces. They are homotopy equivalent if
there exist functions f : X → Y and g : Y → X such that
(i)
f ∘ g : X → X is homotopic to the identity map on X, and
(ii)
g ∘ f : Y → Y is homotopic to the identity map on Y .
If a topological space is homotopy equivalent to a point, then it is said to be contractible.
Question 58.5.4.Why are two homeomorphic spaces also homotopy equivalent?
Intuitively, you can think of this as a more generous form of stretching and bending than
homeomorphism: we are allowed to compress huge spaces into single points.
Example 58.5.5 (ℂ is contractible) Consider the topological spaces ℂ and the space
consisting of the single point {0}. We claim these spaces are homotopy equivalent (can you
guess what f and g are?) Indeed, the two things to check are
(i)
ℂ →{0}ℂ by z00 is homotopy equivalent to the identity on ℂ, which we
just saw, and
(ii)
{0}ℂ →{0} by 000, which is the identity on {0}.
Here by I just mean → in the special case that the function is just an “inclusion”.
Remark 58.5.6 — ℂ cannot be homeomorphic to a point because there is no bijection
of sets between them.
Example 58.5.7 (ℂ ∖{0} is homotopy equivalent to S1) Consider the
topological spaces ℂ ∖{0}, the punctured plane, and the circle S1 viewed as a subset
of S1. We claim these spaces are actually homotopy equivalent! The necessary functions
are the inclusion
and the function
You can check that these satisfy the required condition.
Remark 58.5.8 — On the other hand, ℂ∖{0} cannot be homeomorphic to S1. One can
make S1 disconnected by deleting two points; the same is not true for ℂ ∖{0}.
Example 58.5.9 (Disk = Point, Annulus = Circle.) By the same token, a disk is
homotopic to a point; an annulus is homotopic to a circle. (This might be a little easier
to visualize, since it’s finite.)
I bring these up because it turns out that
Algebraic topology can’t distinguish between homotopy equivalent spaces.
More precisely,
Theorem 58.5.10 (Homotopy equivalent spaces have isomorphic fundamental groups) Let
X and Y be path-connected, homotopy-equivalent spaces. Then πn(X)πn(Y ) for every
positive integer n.
Proof.Let γ : [0,1] → X be a loop. Let f : X → Y and g : Y → X be maps witnessing
that X and Y are homotopy equivalent (meaning f ∘ g and g ∘ f are each homotopic to
the identity). Then the composition
is a loop in Y and hence f induces a natural homomorphism π1(X) → π1(Y ). Similarly
g induces a natural homomorphism π1(Y ) → π1(X). The conditions on f and g now say
exactly that these two homomorphisms are inverse to each other, meaning the maps are
isomorphisms. □
In particular,
Question 58.5.11.What are the fundamental groups of contractible spaces?
That means, for example, that algebraic topology can’t tell the following homotopic
subspaces of ℝ2 apart.
♀♂
58.6The pointed homotopy category
This section is meant to be read by those who know some basic category theory. Those
of you that don’t should come back after reading ?? . Those of you that do will
enjoy how succinctly we can summarize the content of this chapter using categorical
notions.
Definition 58.6.1.The pointed homotopy categoryhTop∗ is defined as follows.
Objects: pairs (X,x0) of spaces with a distinguished basepoint, and
Morphisms: homotopy classes of continuous functions (X,x0) → (Y,y0).
In particular, two path-connected spaces are isomorphic in this category exactly when they are
homotopy equivalent. Then we can summarize many of the preceding results as
follows:
Theorem 58.6.2 (Functorial interpretation of fundamental groups) There is a functor
sending
This implies several things, like
The functor bundles the information of f♯, including the fact that it respects
composition. In the categorical language, f♯ is π1(f).
Homotopic spaces have isomorphic fundamental group (since the spaces are
isomorphic in hTop, and functors preserve isomorphism by ?? ). In fact, you’ll
notice that the proofs of ?? and ?? are secretly identical to each other.
If maps f,g : (X,x0) → (Y,y0) are homotopic, then f♯ = g♯. This is basically ??
Remark 58.6.3 — In fact, π1(X,x0) is the set of arrows (S1,1) → (X,x0) in hTop∗, so
this is actually a covariant Yoneda functor (?? ), except with target Grp instead of Set.
58.7A few harder problems to think about
Problem 58A (Harmonic fan).Exhibit a subspace X of the metric space ℝ2 which is
path-connected but for which a point p can be found such that any r-neighborhood of p
with r < 1 is not path-connected.
Problem 58B† (Special case of Seifert-van Kampen).Let X be a
topological space. Suppose U and V are connected open subsets of X, with X = U ∪V ,
so that U ∩ V is nonempty and path-connected.
Prove that if π1(U) = π1(V ) = {1} then π1(X) = 1.
Remark 58.7.1 — The Seifert–van Kampen theorem generalizes this for π1(U) and
π1(V ) any groups; it gives a formula for calculating π1(X) in terms of π1(U), π1(V ),
π1(U ∩ V ). The proof is much the same.
Unfortunately, this does not give us a way to calculate π1(S1), because it is not possible
to write S1 = U ∪ V for U ∩ V connected.
Problem 58C (RMM 2013).Let n ≥ 2 be a positive integer. A stone
is placed at each vertex of a regular 2n-gon. A move consists of selecting an edge of
the 2n-gon and swapping the two stones at the endpoints of the edge. Prove that if a
sequence of moves swaps every pair of stones exactly once, then there is some edge never
used in any move.
(This last problem doesn’t technically have anything to do with the chapter, but the “gut
feeling” which motivates the solution is very similar.)
59Covering projections
A few chapters ago we talked about what a fundamental group was, but we didn’t actually
show how to compute any of them except for the most trivial case of a simply connected
space. In this chapter we’ll introduce the notion of a covering projection, which will let us see
how some of these groups can be found.
59.1Even coverings and covering projections
Prototypical example for this section:ℝ covers S1.
What we want now is a notion where a big space E, a “covering space”, can be
projected down onto a base space B in a nice way. Here is the notion of “nice”:
Definition 59.1.1.Let p : E → B be a continuous function. Let U be an open set of
B. We call U evenly covered (by p) if ppre(U) is a disjoint union of open sets (possibly
infinite) such that p restricted to any of these sets is a homeomorphism.
Picture:
Image from [?]
All we’re saying is that U is evenly covered if its pre-image is a bunch of copies of it. (Actually, a
little more: each of the pancakes is homeomorphic to U, but we also require that p is the
homeomorphism.)
Definition 59.1.2.A covering projectionp : E → B is a surjective continuous map
such that every base point b ∈ B has an open neighborhood U ∋ b which is evenly
covered by p.
Exercise 59.1.3 (On requiring surjectivity of p).Let p:E → B be satisfying this definition,except that p need not be surjective. Show that the image of p is a connected component of B.Thus if B is connected and E is nonempty, then p:E → B is already surjective. For this reason,some authors omit the surjectivity hypothesis as usually B is path-connected.
Here is the most stupid example of a covering projection.
Example 59.1.4 (Tautological covering projection) Let’s take n disconnected copies of
any space B: formally, E = B ×{1,…,n} with the discrete topology on {1,…,n}. Then
there exists a tautological covering projection E → B by (x,m)x; we just project all
n copies.
This is a covering projection because every open set in B is evenly covered.
This is not really that interesting because B × [n] is not path-connected.
A much more interesting example is that of ℝ and S1.
Example 59.1.5 (Covering projection of S1) Take p : ℝ → S1 by 𝜃e2πi𝜃. This is
essentially wrapping the real line into a single helix and projecting it down.
We claim this is a covering projection. Indeed, consider the point 1 ∈ S1 (where we view S1
as the unit circle in the complex plane). We can draw a small open neighborhood of it whose
pre-image is a bunch of copies in ℝ.
Note that not all open neighborhoods work this time: notably, U = S1 does not work
because the pre-image would be the entire ℝ.
Example 59.1.6 (Covering of S1 by itself) The map S1→ S1 by zz3 is also a
covering projection. Can you see why?
Example 59.1.7 (Covering projections of ℂ ∖{0}) For those comfortable with complex
arithmetic,
(a)
The exponential map exp : ℂ → ℂ ∖{0} is a covering projection.
(b)
For each n, the nth power map −n : ℂ ∖{0}→ ℂ ∖{0} is a covering projection.
59.2Lifting theorem
Prototypical example for this section:ℝ covers S1.
Now here’s the key idea: we are going to try to interpret loops in B as paths in ℝ. This is
often much simpler. For example, we had no idea how to compute the fundamental group of
S1, but the fundamental group of ℝ is just the trivial group. So if we can interpret
loops in S1 as paths in ℝ, that might (and indeed it does!) make computing π1(S1)
tractable.
Definition 59.2.1.Let γ : [0,1] → B be a path and p : E → B a covering projection.
A lifting of γ is a path γ : [0,1] → E such that p ∘γ = γ.
Picture:
Example 59.2.2 (Typical example of lifting) Take p : ℝ → S1⊆ ℂ by 𝜃e2πi𝜃 (so
S1 is considered again as the unit circle). Consider the path γ in S1 which starts at
1 ∈ ℂ and wraps around S1 once, counterclockwise, ending at 1 again. In symbols,
γ : [0,1] → S1 by te2πit.
Then one lifting γ is the path which walks from 0 to 1. In fact, for any integer n, walking
from n to n + 1 works.
Similarly, the counterclockwise path from 1 ∈ S1 to −1 ∈ S1 has a lifting: for some integer n,
the path from n to n + .
The above is the primary example of a lifting. It seems like we have the following structure:
given a path γ in B starting at b0, we start at any point in the fiber ppre(b0). (In our
prototypical example, B = S1, b0 = 1 ∈ ℂ and that’s why we start at any integer n.)
After that we just trace along the path in B, and we get a corresponding path in
E.
Question 59.2.3.Take a path γ in S1with γ(0) = 1 ∈ℂ. Convince yourself that once we selectan integer n ∈ℝ, then there is exactly one lifting starting at n.
It turns out this is true more generally.
Theorem 59.2.4 (Lifting paths) Suppose γ : [0,1] → B is a path with γ(0) = b0, and
p : (E,e0) → (B,b0) is a covering projection. Then there exists a unique lifting
γ : [0,1] → E such that γ(0) = e0.
Proof.For every point b ∈ B, consider an evenly covered open neighborhood Ub in B.
Then the family of open sets
is an open cover of [0,1]. As [0,1] is compact we can take a finite subcover. Thus we
can chop [0,1] into finitely many disjoint closed intervals [0,1] = I1⊔ I2⊔⊔ IN in
that order, such that for every Ik, γimg(Ik) is contained in some Ub.
We’ll construct γ interval by interval now, starting at I1. Initially, place a robot at
e0∈ E and a mouse at b0∈ B. For each interval Ik, the mouse moves around according
to however γ behaves on Ik. But the whole time it’s in some evenly covered Uk; the fact
that p is a covering projection tells us that there are several copies of Uk living in E.
Exactly one of them, say Vk, contains our robot. So the robot just mimics the mouse
until it gets to the end of Ik. Then the mouse is in some new evenly covered Uk+1, and
we can repeat. □
The theorem can be generalized to a diagram
where Y is some general path-connected space, as follows.
Theorem 59.2.5 (General lifting criterion) Let f : (Y,y0) → (B,b0) be continuous
and consider a covering projection p : (E,e0) → (B,b0). (As usual, Y , B, E are
path-connected.) Then a lifting f with f(y0) = e0 exists if and only if
i.e. the image of π1(Y,y0) under f is contained in the image of π1(E,e0) under p (both
viewed as subgroups of π1(B,b0)). If this lifting exists, it is unique.
As p is injective, we actually have p∗img(π1(E,e0))π1(E,e0). But in this case
we are interested in the actual elements, not just the isomorphism classes of the
groups.
Question 59.2.6.What happens if we put Y = [0,1]?
Remark 59.2.7 (Lifting homotopies) — Here’s another cool special case: Recall that a
homotopy can be encoded as a continuous function [0,1] × [0,1] → X. But [0,1] × [0,1]
is also simply connected. Hence given a homotopy γ1≃ γ2 in the base space B, we can
lift it to get a homotopy γ1≃γ2 in E.
Another nice application of this result is ?? .
59.3Lifting correspondence
Prototypical example for this section: (ℝ,0) covers (S1,1).
Let’s return to the task of computing fundamental groups. Consider a covering projection
p : (E,e0) → (B,b0).
A loop γ can be lifted uniquely to γ in E which starts at e0 and ends at some point e in the
fiber ppre(b0). You can easily check that this e ∈ E does not change if we pick a different path
γ′ homotopic to γ.
Question 59.3.1.Look at the picture in?? .
Put one finger at 1 ∈ S1, and one finger on 0 ∈ℝ. Trace a loop homotopic to γ in S1(meaning,you can go backwards and forwards but you must end with exactly one full counterclockwiserotation) and follow along with the other finger in ℝ.
Convince yourself that you have to end at the point 1 ∈ℝ.
Thus every homotopy class of a loop at b0 (i.e. an element of π1(B,b0)) can be
associated with some e in the fiber of b0. The below proposition summarizes this and
more.
Proposition 59.3.2 Let p : (E,e0) → (B,b0) be a covering projection. Then we have a
function of sets
by [γ]γ(1), where γ is the unique lifting starting at e0. Furthermore,
If E is path-connected, then Φ is surjective.
If E is simply connected, then Φ is injective.
Question 59.3.3.Prove that E path-connected implies Φ is surjective. (This is really offensivelyeasy.)
Proof.To prove the proposition, we’ve done everything except show that E simply
connected implies Φ injective. To do this suppose that γ1 and γ2 are loops such that
Φ([γ1]) = Φ([γ2]).
Applying lifting, we get paths γ1 and γ2 both starting at some point e0∈ E and ending
at some point e1∈ E. Since E is simply connected that means they are homotopic, and
we can write a homotopy F : [0,1] × [0,1] → E which unites them. But then consider
the composition of maps
You can check this is a homotopy from γ1 to γ2. Hence [γ1] = [γ2], done. □
This motivates:
Definition 59.3.4.A universal cover of a space B is a covering projection p : E → B
where E is simply connected (and in particular path-connected).
Abuse of Notation 59.3.5.When p is understood, we sometimes just say E is the
universal cover.
Example 59.3.6 (Fundamental group of S1) Let’s return to our standard p : ℝ → S1.
Since ℝ is simply connected, this is a universal cover of S1. And indeed, the fiber of any
point in S1 is a copy of the integers: naturally in bijection with loops in S1.
You can show (and it’s intuitively obvious) that the bijection
is in fact a group homomorphism if we equip ℤ with its additive group structure ℤ. Since
it’s a bijection, this leads us to conclude π1(S1)ℤ.
59.4Regular coverings
Prototypical example for this section:ℝ → S1comes from n ⋅ x = n + x
Here’s another way to generate some coverings. Let X be a topological space and G a group
acting on its points. Thus for every g, we get a map X → X by
We require that this map is continuous1
for every g ∈ G, and that the stabilizer of each point in X is trivial. Then we can consider a
quotient space X∕G defined by fusing any points in the same orbit of this action. Thus the
points of X∕G are identified with the orbits of the action. Then we get a natural
“projection”
by simply sending every point to the orbit it lives in.
Definition 59.4.1.Such a projection is called regular. (Terrible, I know.)
Example 59.4.2 (ℝ → S1 is regular) Let G = ℤ, X = ℝ and define the group action of
G on X by
You can then think of X∕G as “real numbers modulo 1”, with [0,1) a complete set of
representatives and 0 ∼ 1.
So we can identify X∕G with S1 and the associated regular projection is just our usual
exp : 𝜃e2iπ𝜃.
Example 59.4.3 (The torus) Let G = ℤ×ℤ and X = ℝ2, and define the group action of
G on X by (m,n)⋅(x,y) = (m+x,n+y). As [0,1)2 is a complete set of representatives,
you can think of it as a unit square with the edges identified. We obtain the torus S1×S1
and a covering projection ℝ2→ S1× S1.
Example 59.4.4 (ℝℙ2) Let G = ℤ∕2ℤ = and let X = S2 be the surface of
the sphere, viewed as a subset of ℝ3. We’ll let G act on X by sending T ⋅x = −x; hence
the orbits are pairs of opposite points (e.g. North and South pole).
Let’s draw a picture of a space. All the orbits have size two: every point below the
equator gets fused with a point above the equator. As for the points on the equator, we
can take half of them; the other half gets fused with the corresponding antipodes.
Now if we flatten everything, you can think of the result as a disk with half its boundary: this
is ℝℙ2 from before. The resulting space has a name: real projective 2-space, denoted ℝℙ2.
This gives us a covering projection S2→ℝℙ2 (note that the pre-image of a sufficiently small
patch is just two copies of it on S2.)
Example 59.4.5 (Fundamental group of ℝℙ2) As above, we saw that there was a
covering projection S2→ℝℙ2. Moreover the fiber of any point has size two. Since S2 is
simply connected, we have a natural bijection π1(ℝℙ2) to a set of size two; that is,
This can only occur if π1(ℝℙ2)ℤ∕2ℤ, as there is only one group of order two!
Question 59.4.6.Show each of the continuous maps xg ⋅ x is in fact a homeomorphism.(Name its continuous inverse).
59.5The algebra of fundamental groups
Prototypical example for this section:S1, with fundamental group ℤ.
Next up, we’re going to turn functions between spaces into homomorphisms of fundamental
groups.
Let X and Y be topological spaces and f : (X,x0) → (Y,y0). Recall that we defined a group
homomorphism
More importantly, we have:
Proposition 59.5.1 Let p : (E,e0) → (B,b0) be a covering projection of path-connected
spaces. Then the homomorphism p♯ : π1(E,e0) → π1(B,b0) is injective. Hence
p♯img(π1(E,e0)) is an isomorphic copy of π1(E,e0) as a subgroup of π1(B,b0).
Proof.We’ll show kerp♯ is trivial. It suffices to show if γ is a nulhomotopic loop in B
then its lift is nulhomotopic.
By definition, there’s a homotopy F : [0,1] × [0,1] → B taking γ to the constant loop
1B. We can lift it to a homotopy F : [0,1] × [0,1] → E that establishes γ≃1B. But 1E
is a lift of 1B (duh) and lifts are unique. □
Example 59.5.2 (Subgroups of ℤ) Let’s look at the space S1 with fundamental group ℤ. The
group ℤ has two types of subgroups:
The trivial subgroup. This corresponds to the canonical projection ℝ → S1, since
π1(ℝ) is the trivial group (ℝ is simply connected) and hence its image in ℤ is the
trivial group.
nℤ for n ≥ 1. This is given by the covering projection S1→ S1 by zzn. The
image of a loop in the covering S1 is a “multiple of n” in the base S1.
It turns out that these are the only covering projections of Sn by path-connected spaces:
there’s one for each subgroup of ℤ. (We don’t care about disconnected spaces because, again, a
covering projection via disconnected spaces is just a bunch of unrelated “good” coverings.) For
this statement to make sense I need to tell you what it means for two covering projections to
be equivalent.
Definition 59.5.3.Fix a space B. Given two covering projections p1 : E1→ B and
p2 : E2→ B a map of covering projections is a continuous function f : E1→ E2 such that
p2∘ f = p1.
Then two covering projections p1 and p2 are isomorphic if there are f : E1→ E2 and
g : E2→ E1 such that f ∘ g = idE1 and g ∘ f = idE2.
Remark 59.5.4 (For category theorists) — The set of covering projections forms a
category in this way.
It’s an absolute miracle that this is true more generally: the greatest triumph of covering
spaces is the following result. Suppose a space X satisfies some nice conditions, like:
Definition 59.5.5.A space X is called locally connected if for each point x ∈ X and
open neighborhood V of it, there is a connected open set U with x ∈ U ⊆ V .
Definition 59.5.6.A space X is semi-locally simply connected if for every point
x ∈ X there is an open neighborhood U such that all loops in U are nulhomotopic. (But
the contraction need not take place in U.)
Example 59.5.7 (These conditions are weak) Pretty much every space I’ve shown you
has these two properties. In other words, they are rather mild conditions, and you can
think of them as just saying “the space is not too pathological”.
Then we get:
Theorem 59.5.8 (Group theory via covering spaces) Suppose B is a locally connected,
semi-locally simply connected space. Then:
Every subgroup H ⊆ π1(B) corresponds to exactly one covering projection
p : E → B with E path-connected (up to isomorphism).
(Specifically, H is the image of π1(E) in π1(B) through p♯.)
Moreover, the normal subgroups of π1(B) correspond exactly to the regular
covering projections.
Hence it’s possible to understand the group theory of π1(B) completely in terms of the
covering projections.
Moreover, this is how the “universal cover” gets its name: it is the one corresponding to
the trivial subgroup of π1(B). Actually, you can show that it really is universal
in the sense that if p : E → B is another covering projection, then E is in turn
covered by the universal space. More generally, if H1⊆ H2⊆ G are subgroups,
then the space corresponding to H2 can be covered by the space corresponding to
H1.
59.6A few harder problems to think about
Part XVI Category Theory
60Objects and morphisms
I can’t possibly hope to do category theory any justice in these few chapters; thus I’ll just give
a very high-level overview of how many of the concepts we’ve encountered so far can be re-cast
into categorical terms. So I’ll say what a category is, give some examples, then talk about a
few things that categories can do. For my examples, I’ll be drawing from all the previous
chapters; feel free to skip over the examples corresponding to things you haven’t
seen.
If you’re interested in category theory (like I was!), perhaps in what surprising results are
true for general categories, I strongly recommend [?].
60.1Motivation: isomorphisms
From earlier chapters let’s recall the definition of an isomorphism of two objects:
Two groups G and H are isomorphic if there was a bijective homomorphism:
equivalently, we wanted homomorphisms ϕ : G → H and ψ : H → G which were
mutual inverses, meaning ϕ ∘ ψ = idH and ψ ∘ ϕ = idG.
Two metric (or topological) spaces X and Y are isomorphic if there is a continuous
bijection f : X → Y such that f−1 is also continuous.
Two vector spaces V and W are isomorphic if there is a bijection T : V → W
which is a linear map. Again, this can be re-cast as saying that T and T−1 are
linear maps.
Two rings R and S are isomorphic if there is a bijective ring homomorphism ϕ;
again, we can re-cast this as two mutually inverse ring homomorphisms.
In each case we have some collections of objects and some maps, and the isomorphisms
can be viewed as just maps. Let’s use this to motivate the definition of a general
category.
60.2Categories, and examples thereof
Prototypical example for this section:Grpis possibly the most natural example.
Definition 60.2.1.A category𝒜 consists of:
A class of objects, denoted obj(𝒜).
For any two objects A1,A2∈ obj(𝒜), a class of arrows (also called morphisms
or maps) between them. We’ll denote the set of these arrows by Hom𝒜(A1,A2).
For any A1,A2,A3∈ obj(𝒜), if f : A1→ A2 is an arrow and g : A2→ A3 is an
arrow, we can compose these arrows to get an arrow g ∘ f : A1→ A3.
We can represent this in a commutative diagram
where h = g ∘ f. The composition operation ∘ is part of the data of 𝒜; it
must be associative. In the diagram above we say that h factors through
A2.
Finally, every object A ∈ obj(𝒜) has a special identity arrow idA; you can guess what it
does.1
Abuse of Notation 60.2.2.From now on, by A ∈𝒜 we’ll mean A ∈ obj(𝒜).
Abuse of Notation 60.2.3.You can think of “class” as just “set”. The reason we can’t
use the word “set” is because of some paradoxical issues with collections which are too
large; Cantor’s Paradox says there is no set of all sets. So referring to these by “class”
is a way of sidestepping these issues.
Now and forever I’ll be sloppy and assume all my categories are locally small,
meaning that Hom𝒜(A1,A2) is a set for any A1,A2∈𝒜. So elements of 𝒜 may not form
a set, but the set of morphisms between two given objects will always assumed to be a
set.
Let’s formalize the motivation we began with.
Example 60.2.4 (Basic examples of categories)
(a)
There is a category of groups Grp. The data is
The objects of Grp are the groups.
The arrows of Grp are the homomorphisms between these groups.
The composition ∘ in Grp is function composition.
(b)
In the same way we can conceive a category CRing of (commutative) rings.
(c)
Similarly, there is a category Top of topological spaces, whose arrows are the continuous
maps.
(d)
There is a category Top∗ of topological spaces with a distinguished basepoint; that is,
a pair (X,x0) where x0∈ X. Arrows are continuous maps f : X → Y with
f(x0) = y0.
(e)
Similarly, there is a category Vectk of vector spaces (possibly infinite-dimensional) over a
field k, whose arrows are the linear maps. There is even a category FDVectk of
finite-dimensional vector spaces.
(f)
We have a category Set of sets, where the arrows are any maps.
And of course, we can now define what an isomorphism is!
Definition 60.2.5.An arrow A1A2 is an isomorphism if there exists A2A1 such
that f ∘ g = idA2 and g ∘ f = idA1. In that case we say A1 and A2 are isomorphic,
hence A1A2.
Remark 60.2.6 — Note that in Set, XY⇔ = .
Question 60.2.7.Check that every object in a category is isomorphic to itself. (This isoffensively easy.)
More importantly, this definition should strike you as a little impressive. We’re able to
define whether two groups (rings, spaces, etc.) are isomorphic solely by the functions between
the objects. Indeed, one of the key themes in category theory (and even algebra) is
that
One can learn about objects by the functions between them. Category theorytakes this to the extreme by only looking at arrows, and ignoring what theobjects themselves are.
But there are some trickier interesting examples of categories.
Example 60.2.8 (Posets are categories) Let 𝒫 be a partially ordered set. We can construct a
category P for it as follows:
The objects of P are going to be the elements of 𝒫.
The arrows of P are defined as follows:
For every object p ∈ P, we add an identity arrow idp, and
For any pair of distinct objects p ≤ q, we add a single arrow p → q.
There are no other arrows.
There’s only one way to do the composition. What is it?
For example, for the poset 𝒫 on four objects {a,b,c,d} with a ≤ b and a ≤ c ≤ d, we get:
This illustrates the point that
The arrows of a category can be totally different from functions.
In fact, in a way that can be made precise, the term “concrete category” refers to one where
the arrows really are “structure-preserving maps between sets”, like Grp, Top, or
CRing.
Question 60.2.9.Check that no two distinct objects of a poset are isomorphic.
Here’s a second quite important example of a non-concrete category.
Example 60.2.10 (Important: groups are one-Object categories) A group G can be
interpreted as a category 𝒢 with one object ∗, all of whose arrows are isomorphisms.
As [?] says:
The first time you meet the idea that a group is a kind of category, it’s
tempting to dismiss it as a coincidence or a trick. It’s not: there’s real content.
To see this, suppose your education had been shuffled and you took a course
on category theory before ever learning what a group was. Someone comes
to you and says:
“There are these structures called ‘groups’, and the idea is this: a group is
what you get when you collect together all the symmetries of a given thing.”
“What do you mean by a ‘symmetry’?” you ask.
“Well, a symmetry of an object X is a way of transforming X or mapping
X into itself, in an invertible way.”
“Oh,” you reply, “that’s a special case of an idea I’ve met before. A category
is the structure formed by lots of objects and mappings between them – not
necessarily invertible. A group’s just the very special case where you’ve only
got one object, and all the maps happen to be invertible.”
Exercise 60.2.11.Verify the above! That is, show that the data of a one-object category withall isomorphisms is the same as the data of a group.
Finally, here are some examples of categories you can make from other categories.
Example 60.2.12 (Deriving categories)
(a)
Given a category 𝒜, we can construct the opposite category𝒜op, which is the
same as 𝒜 but with all arrows reversed.
(b)
Given categories 𝒜 and ℬ, we can construct the product category𝒜×ℬ as follows:
the objects are pairs (A,B) for A ∈𝒜 and B ∈ℬ, and the arrows from (A1,B1) to
(A2,B2) are pairs
What do you think the composition and identities are?
60.3Special objects in categories
Prototypical example for this section:Sethas initial object ∅ and final object {∗}. An elementof S corresponds to a map {∗}→ S.
Certain objects in categories have special properties. Here are a couple examples.
Example 60.3.1 (Initial object) An initial object of 𝒜 is an object Ainit∈𝒜 such that for
any A ∈𝒜 (possibly A = Ainit), there is exactly one arrow from Ainit to A. For
example,
(a)
The initial object of Set is the empty set ∅.
(b)
The initial object of Grp is the trivial group {1}.
(c)
The initial object of CRing is the ring ℤ (recall that ring homomorphisms R → S
map 1R to 1S).
(d)
The initial object of Top is the empty space.
(e)
The initial object of a partially ordered set is its smallest element, if one exists.
We will usually refer to “the” initial object of a category, since:
Exercise 60.3.2 (Important!).Show that any two initial objects A1, A2of 𝒜 are uniquelyisomorphic meaning there is a unique isomorphism between them.
Remark 60.3.3 — In mathematics, we usually neither know nor care if two objects are
actually equal or whether they are isomorphic. For example, there are many competing
ways to define ℝ, but we still just refer to it as “the” real numbers.
Thus when we define categorical notions, we would like to check they are unique up
to isomorphism. This is really clean in the language of categories, and definitions often
cause objects to be unique up to isomorphism for elegant reasons like the above.
One can take the “dual” notion, a terminal object.
Example 60.3.4 (Terminal object) A terminal object of 𝒜 is an object Afinal∈𝒜 such
that for any A ∈𝒜 (possibly A = Afinal), there is exactly one arrow from A to Afinal. For
example,
(a)
The terminal object of Set is the singleton set {∗}. (There are many singleton sets,
of course, but as sets they are all isomorphic!)
(b)
The terminal object of Grp is the trivial group {1}.
(c)
The terminal object of CRing is the zero ring 0. (Recall that ring homomorphisms
R → S must map 1R to 1S).
(d)
The terminal object of Top is the single-point space.
(e)
The terminal object of a partially ordered set is its maximal element, if one exists.
Again, terminal objects are unique up to isomorphism. The reader is invited to repeat the
proof from the preceding exercise here. However, we can illustrate more strongly the notion of
duality to give a short proof.
Question 60.3.5.Verify that terminal objects of 𝒜 are equivalent to initial objects of 𝒜op. Thusterminal objects of 𝒜 are unique up to isomorphism.
In general, one can consider in this way the dual of any categorical notion: properties of 𝒜
can all be translated to dual properties of 𝒜op (often by adding the prefix “co” in
front).
One last neat construction: suppose we’re working in a concrete category, meaning (loosely)
that the objects are “sets with additional structure”. Now suppose you’re sick of maps
and just want to think about elements of these sets. Well, I won’t let you do that
since you’re reading a category theory chapter, but I will offer you some advice:
In Set, arrows from {∗} to S correspond to elements of S.
In Top, arrows from {∗} to X correspond to points of X.
In Grp, arrows from ℤ to G correspond to elements of G.
In CRing, arrows from ℤ[x] to R correspond to elements of R.
and so on. So in most concrete categories, you can think of elements as functions from special
sets to the set in question. In each of these cases we call the object in question a freeobject.
60.4Binary products
Prototypical example for this section:X × Y in most concrete categories is the set-theoreticproduct.
The “universal property” is a way of describing objects in terms of maps in such a way that
it defines the object up to unique isomorphism (much the same as the initial and terminal
objects).
To show how this works in general, let me give a concrete example. Suppose I’m in a
category – let’s say Set for now. I have two sets X and Y , and I want to construct the
Cartesian product X × Y as we know it. The philosophy of category theory dictates that I
should talk about maps only, and avoid referring to anything about the sets themselves. How
might I do this?
Well, let’s think about maps into X × Y . The key observation is that
A function AX × Y amounts to a pair of functions (AX,AY ).
Put another way, there is a natural projection map X × Y ↠ X and X × Y ↠ Y :
(We have to do this in terms of projection maps rather than elements, because category theory
forces us to talk about arrows.) Now how do I add A to this diagram? The point is that there
is a bijection between functions AX × Y and pairs (g,h) of functions. Thus for every pair
AX and AY there is a unique function AX × Y .
But X ×Y is special in that it is “universal”: for any other set A, if you give me functions
A → X and A → Y , I can use it build a unique function A → X × Y . Picture:
We can do this in any general category, defining a so-called product.
Definition 60.4.1.Let X and Y be objects in any category 𝒜. The product consists of an
object X ×Y and arrows πX, πY to X and Y (thought of as projection). We require that for
any object A and arrows AX, AY , there is a unique function AX × Y such that the
diagram
commutes.
Abuse of Notation 60.4.2.Strictly speaking, the product should consist of both the
object X×Y and the projection maps πX and πY. However, if πX and πY are understood,
then we often use X × Y to refer to the object, and refer to it also as the product.
Products do not always exist; for example, take a category with just two objects and no
non-identity morphisms. Nonetheless:
Proposition 60.4.3 (Uniqueness of products) When they exist, products are unique up to
isomorphism: given two products P1 and P2 of X and Y there is an isomorphism between
the two objects.
Proof.This is very similar to the proof that initial objects are unique up to unique
isomorphism. Consider two such objects P1 and P2, and the associated projection maps. So,
we have a diagram
There are unique morphisms f and g between P1 and P2 that make the entire diagram
commute, according to the universal property.
On the other hand, look at g ∘f and focus on just the outer square. Observe that g ∘f is a
map which makes the outer square commute, so by the universal property of P1 it is the only
one. But idP1 works as well. Thus idP1 = g ∘ f. Similarly, f ∘ g = idP2 so f and g are
isomorphisms. □
Abuse of Notation 60.4.4.Actually, this is not really the morally correct theorem;
since we’ve only showed the objects P1 and P2 are isomorphic and have not made any
assertion about the projection maps. But I haven’t (and won’t) define isomorphism of
the entire product, and so in what follows if I say “P1 and P2 are isomorphic” I really
just mean the objects are isomorphic.
Exercise 60.4.5.In fact, show the products are unique up to unique isomorphism: the f and gabove are the only isomorphisms between the objects P1and P2.
The nice fact about this “universal property” mindset is that we don’t have to give explicit
constructions; assuming existence, the “universal property” allows us to bypass all this work
by saying “the object with these properties is unique up to unique isomorphism”,
thus we don’t need to understand the internal workings of the object to use its
properties.
Of course, that’s not to say we can’t give concrete examples.
Example 60.4.6 (Examples of products)
(a)
In Set, the product of two sets X and Y is their Cartesian product X × Y .
(b)
In Grp, the product of G, H is the group product G × H.
(c)
In Vectk, the product of V and W is V ⊕ W.
(d)
In CRing, the product of R and S is appropriately the ring product R × S.
(e)
Let 𝒫 be a poset interpreted as a category. Then the product of two objects x and y is the
greatest lower bound; for example,
If the poset is (ℝ,≤) then it’s min{x,y}.
If the poset is the subsets of a finite set by inclusion, then it’s x ∩ y.
If the poset is the positive integers ordered by division, then it’s gcd(x,y).
Of course, we can define products of more than just one object. Consider a set of objects
(Xi)i∈I in a category 𝒜. We define a cone on the Xi to be an object A with some
“projection” maps to each Xi. Then the product is a cone P which is “universal” in the same
sense as before: given any other cone A there is a unique map A → P making the diagram
commute. In short, a product is a “universal cone”.
The picture of this is
See also ?? .
One can also do the dual construction to get a coproduct: given X and Y , it’s the object
X + Y together with maps XX + Y and Y X + Y (that’s Greek iota, think inclusion)
such that for any object A and maps XA, Y A there is a unique f for which
commutes. We’ll leave some of the concrete examples as an exercise this time, for
example:
Exercise 60.4.7.Describe the coproduct inSet.
Predictable terminology: a coproduct is a universal cocone.
Spoiler alert later on: this construction can be generalized vastly to so-called “limits”, and
we’ll do so later on.
60.5Monic and epic maps
The notion of “injective” doesn’t make sense in an arbitrary category since arrows need not be
functions. The correct categorical notion is:
Definition 60.5.1.A map XY is monic (or a monomorphism) if for any commutative
diagram
we must have g = h. In other words, f ∘ g = f ∘ hg = h.
Question 60.5.2.Verify that in a concrete category, injectivemonic.
Question 60.5.3.Show that the composition of two monic maps is monic.
In most but not all situations, the converse is also true.
Exercise 60.5.4.Show that inSet,Grp,CRing, monic implies injective. (Take A = {∗}, A = ℤ,A = ℤ[x].)
More generally, as we said before there are many categories with a “free” object that you
can use to think of as elements. An element of a set is a function 1 → S, and element of a ring
is a function ℤ[x] → R, et cetera. In all these categories, the definition of monic literally reads
“f is injective on Hom𝒜(A,X)”. So in these categories, “monic” and “injective”
coincide.
That said, here is the standard counterexample. An additive abelian group G = (G,+) is
called divisible if for every x ∈ G and n ∈ ℤ there exists y ∈ G with ny = x. Let DivAbGrp be
the category of such groups.
Exercise 60.5.5.Show that the projection ℚ →ℚ∕ℤ is monic but not injective.
Of course, we can also take the dual notion.
Definition 60.5.6.A map XY is epic (or an epimorphism) if for any commutative
diagram
we must have g = h. In other words, g ∘ f = h ∘ fg = h.
This is kind of like surjectivity, although it’s a little farther than last time. Note that in
concrete categories, surjective epic.
Exercise 60.5.7.Show that inSet,Grp,Ab,Vectk,Top, the notions of epic and surjectivecoincide. (ForSet, take A = {0,1}.)
However, there are more cases where it fails. Most notably:
Example 60.5.8 (Epic but not surjective)
(a)
In CRing, for instance, the inclusion ℤℚ is epic (and not surjective).. Indeed,
if two homomorphisms ℚ → A agree on every integer then they agree everywhere
(why?),
(b)
In the category of Hausdorff topological spaces (every two points have disjoint open
neighborhoods), in fact epic ⇔ dense image (like ℚℝ).
Thus failures arise when a function f : X → Y can be determined by just some of the points of
X.
60.6A few harder problems to think about
Problem 60A.In the category Vectk of k-vector spaces (for a field k), what are the
initial and terminal objects?
Problem 60B†.What is the coproduct X+Y in the categories Set, Vectk, and a poset?
Problem 60C.In any category 𝒜 where all products exist, show that
where X, Y , Z are arbitrary objects. (Here both sides refer to the objects, as in ?? .)
Problem 60D.Consider a category 𝒜 with a zero object, meaning
an object which is both initial and terminal. Given objects X and Y in A, prove that
the projection X × Y → X is epic.
61Functors and natural transformations
Functors are maps between categories; natural transformations are maps between
functors.
61.1Many examples of functors
Prototypical example for this section:Forgetful functors; fundamental groups; −∨.
Here’s the point of a functor:
Pretty much any time you make an object out of another object, you get afunctor.
Before I give you a formal definition, let me list (informally) some examples. (You’ll notice
some of them have opposite categories 𝒜op appearing in places. Don’t worry about those for
now; you’ll see why in a moment.)
Given a group G (or vector space, field, …), we can take its underlying set S; this
is a functor from Grp→Set.
Given a set S we can consider a vector space with basis S; this is a functor from
Set→Vect.
Given a vector space V we can consider its dual space V∨. This is a functor
Vectkop→Vectk.
Tensor products give a functor from Vectk×Vectk→Vectk.
Given a set S, we can build its power set, giving a functor Set→Set.
In algebraic topology, we take a topological space X and build several groups
H1(X), π1(X), etc. associated to it. All these group constructions are functors
Top→Grp.
Sets of homomorphisms: let 𝒜 be a category.
Given two vector spaces V1 and V2 over k, we construct the abelian group of
linear maps V1→ V2. This is a functor from Vectkop×Vectk→AbGrp.
More generally for any category 𝒜 we can take pairs (A1,A2) of objects and
obtain a set Hom𝒜(A1,A2). This turns out to be a functor 𝒜op×𝒜→Set.
The above operation has two “slots”. If we “pre-fill” the first slots, then we
get a functor 𝒜→Set. That is, by fixing A ∈𝒜, we obtain a functor (called
HA) from 𝒜 →Set by sending A′ ∈ 𝒜 to Hom𝒜(A,A′). This is called the
covariant Yoneda functor (explained later).
As we saw above, for every A ∈ 𝒜 we obtain a functor HA : 𝒜 →Set. It
turns out we can construct a category [𝒜,Set] whose elements are functors
𝒜→Set; in that case, we now have a functor 𝒜op→ [𝒜,Set].
61.2Covariant functors
Prototypical example for this section:Forgetful/free functors, …
Category theorists are always asking “what are the maps?”, and so we can now think about
maps between categories.
Definition 61.2.1.Let 𝒜 and ℬ be categories. Of course, a functorF takes every object of
𝒜 to an object of ℬ. In addition, though, it must take every arrow A1A2 to an arrow
F(A1)F(A2). You can picture this as follows.
(I’ll try to use dotted arrows for functors, which cross different categories, for emphasis.) It
needs to satisfy the “naturality” requirements:
Identity arrows get sent to identity arrows: for each identity arrow idA, we have
F(idA) = idF(A).
The functor respects composition: if A1A2A3 are arrows in 𝒜, then F(g∘f) =
F(g) ∘ F(f).
So the idea is:
Whenever we naturally make an object A ∈𝒜 into an object of B ∈ℬ, thereshould usually be a natural way to transform a map A1→ A2into a mapB1→ B2.
Let’s see some examples of this.
Example 61.2.2 (Free and forgetful functors) Note that these are both informal terms, and
don’t have a rigid definition.
(a)
We talked about a forgetful functor earlier, which takes the underlying set of a
category like Vectk. Let’s call it U : Vectk→Set.
Now, given a map T : V1→ V2 in Vectk, there is an obvious U(T) : U(V1) → U(V2)
which is just the set-theoretic map corresponding to T.
Similarly there are forgetful functors from Grp, CRing, etc., to Set. There is even
a forgetful functor CRing→Grp: send a ring R to the abelian group (R,+). The
common theme is that we are “forgetting” structure from the original category.
(b)
We also talked about a free functor in the example. A free functor F : Set→Vectk
can be taken by considering F(S) to be the vector space with basis S. Now, given
a map f : S → T, what is the obvious map F(S) → F(T)? Simple: take each basis
element s ∈ S to the basis element f(s) ∈ T.
Similarly, we can define F : Set→Grp by taking the free group generated by a set
S.
Remark 61.2.3 — There is also a notion of “injective” and “surjective” for functors
(on arrows) as follows. A functor F : 𝒜→ℬ is faithful (resp. full) if for any A1,A2,
F : Hom𝒜(A1,A2) → Homℬ(FA1,FA2) is injective (resp. surjective).a
We can use this to give an exact definition of concrete category: it’s a category with a
faithful (forgetful) functor U : 𝒜→Set.
Example 61.2.4 (Functors from 𝒢) Let G be a group and 𝒢 = {∗} be the associated
one-object category.
(a)
Consider a functor F : 𝒢→Set, and let S = F(∗). Then the data of F corresponds
to putting a group action of G on S.
(b)
Consider a functor F : 𝒢→FDVectk, and let V = F(∗) have dimension n. Then the
data of F corresponds to embedding G as a subgroup of the n×n matrices (i.e. the
linear maps V → V ). This is one way groups historically arose; the theory of viewing
groups as matrices forms the field of representation theory.
(c)
Let H be a group and construct ℋ the same way. Then functors 𝒢→ℋ correspond
to homomorphisms G → H.
Exercise 61.2.5.Check the above group-based functors work as advertised.
Here’s a more involved example. If you find it confusing, skip it and come back after reading
about its contravariant version.
Example 61.2.6 (Covariant Yoneda functor) Fix an A ∈𝒜. For a category 𝒜, define
the covariant Yoneda functorHA: 𝒜→Set by defining
Hence each A1 is sent to the arrows from A to A1; so HAdescribes how A sees theworld.
Now we want to specify how HA behaves on arrows. For each arrow A1A2, we need
to specify Set-map Hom𝒜(A,A1) → Hom(A,A2); in other words, we need to send an
arrow AA1 to an arrow A → A2. There’s only one reasonable way to do this: take the
composition
In other words, HA(f) is pf ∘ p. In still other words, HA(f) = f ∘−; the − is a slot
for the input to go into.
As another example:
Question 61.2.7.If 𝒫 and 𝒬 are posets interpreted as categories, what does a functor from 𝒫to 𝒬 represent?
Now, let me explain why we might care. Consider the following “obvious” fact: if G and H
are isomorphic groups, then they have the same size. We can formalize it by saying: if
GH in Grp and U : Grp→Set is the forgetful functor (mapping each group to its
underlying set), then U(G)U(H). The beauty of category theory shows itself: this in
fact works for any functors and categories, and the proof is done solely through
arrows:
Theorem 61.2.8 (Functors preserve isomorphism) If A1A2 are isomorphic objects in 𝒜
and F : 𝒜→ℬ is a functor then F(A1)F(A2).
Proof.Try it yourself! The picture is:
You’ll need to use both key properties of functors: they preserve composition and the identity
map. □
This will give us a great intuition in the future, because
(i)
Almost every operation we do in our lifetime will be a functor, and
(ii)
We now know that functors take isomorphic objects to isomorphic objects.
Thus, we now automatically know that basically any “reasonable” operation we do will preserve
isomorphism (where “reasonable” means that it’s a functor). This is super convenient in
algebraic topology, for example; see ?? , where we get for free that homotopic spaces have
isomorphic fundamental groups.
Remark 61.2.9 — This lets us construct a category Cat whose objects are categories
and arrows are functors.
61.3Contravariant functors
Prototypical example for this section:Dual spaces, contravariant Yoneda functor,etc.
Now I have to explain what the opposite categories were doing earlier. In all the previous
examples, we took an arrow A1→ A2, and it became an arrow F(A1) → F(A2). Sometimes,
however, the arrow in fact goes the other way: we get an arrow F(A2) → F(A1) instead. In
other words, instead of just getting a functor 𝒜→ℬ we ended up with a functor
𝒜op→ℬ.
These functors have a name:
Definition 61.3.1.A contravariant functor from 𝒜 to ℬ is a functor F : 𝒜op→ℬ.
(Note that we do not write “contravariant functor F : 𝒜 → ℬ”, since that would be
confusing; the function notation will always use the correct domain and codomain.)
Pictorially:
For emphasis, a usual functor is often called a covariant functor. (The word “functor” with
no adjective always refers to covariant.)
Let’s see why this might happen.
Example 61.3.2 (VV∨ is contravariant) Consider the functor Vectk→Vectk by
VV∨.
If we were trying to specify a covariant functor, we would need, for every linear map
T : V1→ V2, a linear map T∨ : V 1∨→ V 2∨. But recall that V 1∨ = Hom(V1,k) and
V 2∨ = Hom(V2,k): there’s no easy way to get an obvious map from left to right.
However, there is an obvious map from right to left: given ξ2 : V2→ k, we can easily give a
map from V1→ k: just compose with T! In other words, there is a very natural map
V 2∨→ V 1∨ according to the composition
In summary, a map T : V1→ V2 induces naturally a map T∨ : V 2∨→ V 1∨ in the opposite
direction. So the contravariant functor looks like:
We can generalize the example above in any category by replacing the field k with any
chosen object A ∈𝒜.
Example 61.3.3 (Contravariant Yoneda functor) The contravariantYoneda functor on 𝒜, denoted HA : 𝒜op→Set, is used to describe how objects of 𝒜
see A. For each X ∈𝒜 it puts
For XY in 𝒜, the map HA(f) sends each arrow Y A ∈ Hom𝒜(Y,A) to
as we did above. Thus HA(f) is an arrow from Hom𝒜(Y,A) → Hom𝒜(X,A). (Note the
flipping!)
Exercise 61.3.4.Check now the claim that 𝒜op×𝒜→Setby (A1,A2)Hom(A1,A2) is infact a functor.
61.4Equivalence of categories
61.5(Optional) Natural transformations
We made categories to keep track of objects and maps, then went a little crazy and asked
“what are the maps between categories?” to get functors. Now we’ll ask “what are the maps
between functors?” to get natural transformations.
It might sound terrifying that we’re drawing arrows between functors, but this is actually an
old idea. Recall that given two paths α,β : [0,1] → X, we built a path-homotopy
by “continuously deforming” the path α to β; this could be viewed as a function
[0,1] × [0,1] → X. The definition of a natural transformation is similar: we want to pull F to
G along a series of arrows in the target space ℬ.
Definition 61.5.1.Let F,G : 𝒜→ℬ be two functors. A natural transformationα
from F to G, denoted
consists of, for each A ∈𝒜 an arrow αA∈ Homℬ(F(A),G(A)), which is called the
component of α at A. Pictorially, it looks like this:
These αA are subject to the “naturality” requirement that for any A1A2, the diagram
commutes.
The arrow αA represents the path that F(A) takes to get to G(A) (just as in a path-homotopy
from α to β each point α(t) gets deformed to the point β(t) continuously). A picture might
help: consider
Here 𝒜 is the small category with three elements and two non-identity arrows f, g (I’ve omitted
the identity arrows for simplicity). The images of 𝒜 under F and G are the blue and
green “subcategories” of ℬ. Note that ℬ could potentially have many more objects
and arrows in it (grey). The natural transformation α (red) selects an arrow of
ℬ from each F(A) to the corresponding G(A), dragging the entire image of F to
the image of G. Finally, we require that any diagram formed by the blue, red, and
green arrows is commutative (naturality), so the natural transformation is really
“natural”.
There is a second equivalent definition that looks much more like the homotopy.
Definition 61.5.2.Let 2 denote the category generated by a poset with two elements 0 ≤ 1,
that is,
Then a natural transformation is just a functor α : 𝒜×2 →ℬ satisfying
More succinctly, α(−,0) = F, α(−,1) = G.
The proof that these are equivalent is left as a practice problem.
Naturally, two natural transformations α : F → G and β : G → H can get composed.
Now suppose α is a natural transformation such that αA is an isomorphism for each A. In
this way, we can construct an inverse arrow βA to it.
In this case, we say α is a natural isomorphism. We can then say that F(A)G(A)
naturally in A. (And β is an isomorphism too!) This means that the functors F and G are
“really the same”: not only are they isomorphic on the level of objects, but these isomorphisms
are “natural”. As a result of this, we also write FG to mean that the functors are naturally
isomorphic.
This is what it really means when we say that “there is a natural / canonical isomorphism”.
For example, I claimed earlier (in ?? ) that there was a canonical isomorphism (V∨)∨V ,
and mumbled something about “not having to pick a basis” and “God-given”. Category
theory, amazingly, lets us formalize this: it just says that (V∨)∨id(V ) naturally in
V ∈FDVectk. Really, we have a natural transformation
where the component 𝜀V is given by vev v (as discussed earlier, the fact that it is an
isomorphism follows from the fact that V and (V∨)∨ have equal dimensions and 𝜀V is
injective).
61.6(Optional) The Yoneda lemma
Now that I have natural transformations, I can define:
Definition 61.6.1.The functor category of two categories 𝒜 and ℬ, denoted [𝒜,ℬ], is
defined as follows:
The objects of [𝒜,ℬ] are (covariant) functors F : 𝒜→ℬ, and
The morphisms are natural transformations α : F → G.
Question 61.6.2.When are two objects in the functor category isomorphic?
With this, I can make good on the last example I mentioned at the beginning:
Exercise 61.6.3.Construct the following functors:
𝒜→ [𝒜op,Set] by AHA, which we call H∙.
𝒜op→ [𝒜,Set] by AHA, which we call H∙.
Notice that we have opposite categories either way; even if you like HA because it
is covariant, the map H∙ is contravariant. So for what follows, we’ll prefer to use
H∙.
The main observation now is that given a category 𝒜, H∙ provides some special functors
𝒜op→Set which are already “built” in to the category A. In light of this, we define:
Definition 61.6.4.A presheafX is just a contravariant functor 𝒜op→Set. It is called
representable if XHA for some A.
In other words, when we think about representable, the question we’re asking is:
What kind of presheaves are already “built in” to the category 𝒜?
One way to get at this question is: given a presheaf X and a particular HA, we can look at the
set of natural transformations α : XHA, and see if we can learn anything about it. In
fact, this set can be written explicitly:
Theorem 61.6.5 (Yoneda lemma) Let 𝒜 be a category, pick A ∈𝒜, and let HA be the
contravariant Yoneda functor. Let X : 𝒜op→Set be a contravariant functor. Then the
map
defined by ααA(idA) ∈ X(A) is an isomorphism of Set (i.e. a bijection). Moreover, if
we view both sides of the equality as functors
then this isomorphism is natural.
This might be startling at first sight. Here’s an unsatisfying explanation why this might not
be too crazy: in category theory, a rule of thumb is that “two objects of the same type that are
built naturally are probably the same”. You can see this theme when we defined functors and
natural transformations, and even just compositions. Now to look at the set of natural
transformations, we took a pair of elements A ∈𝒜 and X ∈ [𝒜op,Set] and constructed a set of
natural transformations. Is there another way we can get a set from these two pieces of
information? Yes: just look at X(A). The Yoneda lemma is telling us that our heuristic still
holds true here.
Some consequences of the Yoneda lemma are recorded in [?]. Since this chapter is
already a bit too long, I’ll just write down the statements, and refer you to [?] for the
proofs.
1.
As we mentioned before, H∙ provides a functor
It turns out this functor is in fact fully faithful; it quite literally embeds the category
𝒜 into the functor category on the right (much like Cayley’s theorem embeds every
group into a permutation group).
2.
If X,Y ∈𝒜 then
To see why this is expected, consider 𝒜 = Grp for concreteness. Suppose A, X, Y are
groups such that HX(A)HY(A) for all A. For example,
If A = ℤ, then = .
If A = ℤ∕2ℤ, then X and Y have the same number of elements of order 2.
…
Each A gives us some information on how X and Y are similar, but the whole natural
isomorphism is strong enough to imply XY .
3.
Consider the functor U : Grp→Set. It can be represented by Hℤ, in the sense
that
That is, elements of G are in bijection with maps ℤ → G, determined by the image of +1
(or −1 if you prefer). So a representation of U was determined by looking at ℤ and
picking +1 ∈ U(ℤ).
The generalization of this is a follows: let 𝒜 be a category and X : 𝒜→Set a covariant
functor. Then a representation HAX consists of an object A ∈𝒜 and an
element u ∈ X(A) satisfying a certain condition. You can read this off the
condition11 Just for completeness, the condition is: For all A′∈𝒜and x ∈X(A′), there’s a unique f : A →A′with(Xf)(u) = x.
if you know what the inverse map is in ?? . In the above situation, X = U, A = ℤ and
u = ±1.
61.7A few harder problems to think about
Problem 61A.Show that the two definitions of natural transformation (one in
terms of 𝒜×2 →ℬ and one in terms of arrows F(A)G(A)) are equivalent.
Problem 61B.Let 𝒜 be the category of finite sets whose arrows are bijections between sets.
For A ∈𝒜, let F(A) be the set of permutations of A and let G(A) be the set of orderings on
A.2
(a)
Extend F and G to functors 𝒜→Set.
(b)
Show that F(A)G(A) for every A, but this isomorphism is not natural.
Problem 61C (Proving the Yoneda lemma).In the context of ?? :
(a)
Prove that the map described is in fact a bijection. (To do this, you will probably
have to explicitly write down the inverse map.)
(b)
Prove that the bijection is indeed natural. (This is long-winded,
but not difficult; from start to finish, there is only one thing you can possibly do.)
62Limits in categories (TO DO)
We saw near the start of our category theory chapter the nice construction of products by
drawing a bunch of arrows. It turns out that this concept can be generalized immensely, and I
want to give a you taste of that here.
To run this chapter, we follow the approach of [?].
62.1Equalizers
Prototypical example for this section:The equalizer of f,g : X → Y is the set of points withf(x) = g(x).
Given two sets X and Y , and maps XY , we define their equalizer to be
We would like a categorical way of defining this, too.
Consider two objects X and Y with two maps f and g between them. Stealing a page from
[?], we call this a fork:
A cone over this fork is an object A and arrows over X and Y which make the diagram
commute, like so.
Effectively, the arrow over Y is just forcing f ∘q = g ∘q. In any case, the equalizer of f and g
is a “universal cone” over this fork: it is an object E and a map EX such that for each
AX the diagram
commutes for a unique AE. In other words, any map AX as above must factor uniquely
through E. Again, the dotted arrows can be omitted, and as before equalizers may not exist.
But when they do exist:
Exercise 62.1.1.If EX and E′X are equalizers, show that EE′.
Example 62.1.2 (Examples of equalizers)
(a)
In Set, given XY the equalizer E can be realized as E = {x∣f(x) = g(x)}, with
the inclusion e : EX as the morphism. As usual, by abuse we’ll often just refer
to E as the equalizer.
(b)
Ditto in Top, Grp. One has to check that the appropriate structures are preserved
(e.g. one should check that {ϕ(g) = ψ(g)∣g ∈ G} is a group).
(c)
In particular, given a homomorphism ϕ : G → H, the inclusion kerϕG is an
equalizer for the fork G → H by ϕ and the trivial homomorphism.
According to (c) equalizers let us get at the concept of a kernel if there is a distinguished
“trivial map”, like the trivial homomorphism in Grp. We’ll flesh this idea out in the chapter on
abelian categories.
62.2Pullback squares (TO DO)
Great example: differentiable functions on (−3,1) and (−1,3)
Example 62.2.1
62.3Limits
We’ve defined cones over discrete sets of Xi and over forks. It turns out you can also define a
cone over any general diagram of objects and arrows; we specify a projection from A to each
object and require that the projections from A commute with the arrows in the diagram. (For
example, a cone over a fork is a diagram with two edges and two arrows.) If you then demand
the cone be universal, you have the extremely general definition of a limit. As always, these
are unique up to unique isomorphism. We can also define the dual notion of a colimit in the
same way.
62.4A few harder problems to think about
Problem 62A⋆ (Equalizers are monic).Show that the equalizer of any fork is monic.
pushout square gives tenor product
p-adic
relative Chinese remainder theorem!!
63Abelian categories
In this chapter I’ll translate some more familiar concepts into categorical language; this will
require some additional assumptions about our category, culminating in the definition of a
so-called “abelian category”. Once that’s done, I’ll be able to tell you what this “diagram
chasing” thing is all about.
Throughout this chapter, “” will be used for monic maps and “↠” for epic
maps.
63.1Zero objects, kernels, cokernels, and images
Prototypical example for this section:InGrp, the trivial group and homomorphism are thezero objects and morphisms. If G, H are abelian then the cokernel of ϕ : G → H isH∕imϕ.
A zero object of a category is an object 0 which is both initial and terminal; of course, it’s
unique up to unique isomorphism. For example, in Grp the zero object is the trivial
group, in Vectk it’s the zero-dimensional vector space consisting of one point, and so
on.
Question 63.1.1.Show thatSetandTopdon’t have zero objects.
For the rest of this chapter, all categories will have zero objects.
In a category 𝒜 with zero objects, any two objects A and B thus have a distinguished
morphism
which is called the zero morphism and also denoted 0. For example, in Grp this is the
trivial homomorphism.
We can now define:
Definition 63.1.2.Consider a map AB. The kernel is defined as the equalizer of this map
and the map AB. Thus, it’s a map kerf : KerfA such that
commutes, and moreover any other map with the same property factors uniquely through
KerA (so it is universal with this property). By ?? , kerf is a monic morphism, which
justifies the use of “”.
Notice that we’re using kerf to represent the map and Kerf to represent the object Similarly,
we define the cokernel, the dual notion:
Definition 63.1.3.Consider a map AB. The cokernel of f is a map cokerf : B ↠ Cokerf
such that
commutes, and moreover any other map with the same property factors uniquely through
Cokerf (so it is universal with this property). Thus it is the “coequalizer” of this map and the
map AB. By the dual of ?? , cokerf is an epic morphism, which justifies the use of
“↠”.
Think of the cokernel of a map AB as “B modulo the image of f”, e.g.
Example 63.1.4 (Cokernels) Consider the map
ℤ∕6ℤ → D12 = . Then the cokernel of this map in Grp is
D12∕ℤ∕2ℤ.
This doesn’t always work out quite the way we want since in general the image of a
homomorphism need not be normal in the codomain. Nonetheless, we can use this to define:
Definition 63.1.5.The image of AB is the kernel of cokerf. We denote Imf =
Ker(cokerf). This gives a unique map imf : A → Imf.
When it exists, this coincides with our concrete notion of “image”. Picture:
Note that by universality of Imf, we find that there is a unique map imf : A → Imf that
makes the entire diagram commute.
63.2Additive and abelian categories
Prototypical example for this section:Ab,Vectk, or more generallyModR.
We can now define the notion of an additive and abelian category, which are the types of
categories where this notion is most useful.
Definition 63.2.1.An additive category𝒜 is one such that:
𝒜 has a zero object, and any two objects have a product.
More importantly: every Hom𝒜(A,B) forms an abelian group (written additively)
such that composition distributes over addition:
The zero map serves as the identity element for each group.
Definition 63.2.2.An abelian category𝒜 is one with the additional properties that for
any morphism AB,
The kernel and cokernel exist, and
The morphism factors through the image so that im(f) is epic.
So, this yields a diagram
Example 63.2.3 (Examples of abelian categories)
(a)
Vectk, Ab are abelian categories, where f + g takes its usual meaning.
(b)
Generalizing this, the category ModR of R-modules is abelian.
(c)
Grp is not even additive, because there is no way to assign a commutative addition
to pairs of morphisms.
In general, once you assume a category is abelian, all the properties you would want of these
kernels, cokernels, … that you would guess hold true. For example,
Proposition 63.2.4 (Monic⇔trival kernel) A map AB is monic if and only if its kernel
is 0 → A. Dually, AB is epic if and only if its cokernel is B → 0.
Proof.The easy direction is:
Exercise 63.2.5.Show that if AB is monic, then 0 → A is a kernel. (This holds even innon-abelian categories.)
Of course, since kernels are unique up to isomorphism, monic 0 kernel. On
the other hand, assume that 0 → A is a kernel of AB. For this we can exploit
the group structure of the underlying homomorphisms now. Assume the diagram
commutes. Then (g −h) ∘f = g ∘f −h∘f = 0, and we’ve arrived at a commutative diagram.
But since 0 → A is a kernel it follows that g − h factors through 0, so g − h = 0 g = h,
which is to say that f is monic. □
Proposition 63.2.6 (Isomorphism⇔monic and epic) In an abelian category, a map is an
isomorphism if and only if it is monic and epic.
Proof.Omitted. (The Mitchell embedding theorem presented later implies this anyways
for most situations we care about, by looking at a small sub-category.) □
63.3Exact sequences
Prototypical example for this section: 0 → G → G × H → H → 0 is exact.
Exact sequences will seem exceedingly unmotivated until you learn about homology groups,
which is one of the most natural places that exact sequences appear. In light of this, it might
be worth trying to read the chapter on homology groups simultaneously with this
one.
First, let me state the definition for groups, to motivate the general categorical definition. A
sequence of groups
is exact at Gk if the image of fk is the kernel of fk+1. We say the entire sequence is exact if
it’s exact at k = 1,…,n − 1.
Example 63.3.1 (Exact sequences)
(a)
The sequence
is exact. Actually, 0 → GG × H ↠ H → 0 is exact in general. (Here 0 denotes
the trivial group.)
(b)
For groups, the map 0 → A → B is exact if and only if A → B is injective.
(c)
For groups, the map A → B → 0 is exact if and only if A → B is surjective.
Now, we want to mimic this definition in a general abelian category 𝒜. So, let’s write down
a criterion for when ABC is exact. First, we had better have that g ∘ f = 0, which
encodes the fact that im(f) ⊆ ker(g). Adding in all the relevant objects, we get the
commutative diagram below.
Here the map A ↠ Imf is epic since we are assuming 𝒜 is an abelian category. So, we have
that
but since imf is epic, this means that g ∘ι = 0. So there is a unique map Imf → Kerg, and
we require that this diagram commutes. In short,
Definition 63.3.2.Let 𝒜 be an abelian category. The sequence
is exact at An if fn∘ fn+1 = 0 and the canonical map Imfn→ Kerfn+1 is an
isomorphism. The entire sequence is exact if it is exact at each Ai. (For finite sequences
we don’t impose condition on the very first and very last object.)
Exercise 63.3.3.Show that, as before, 0 → A → B is exact⇔A → B is monic.
63.4The Freyd-Mitchell embedding theorem
We now introduce the Freyd-Mitchell embedding theorem, which essentially says that any
abelian category can be realized as a concrete one.
Definition 63.4.1.A category is small if obj(𝒜) is a set (as opposed to a class),
i.e. there is a “set of all objects in 𝒜”. For example, Set is not small because there is no
set of all sets.
Theorem 63.4.2 (Freyd-Mitchell embedding theorem) Let 𝒜 be a small abelian category.
Then there exists a ring R (with 1 but possibly non-commutative) and a full, faithful,
exact functor onto the category of left R-modules.
Here a functor is exact if it preserves exact sequences. This theorem is good because it
means
You can basically forget about all the weird definitions that work in any abeliancategory.
Any time you’re faced with a statement about an abelian category, it suffices to just prove it
for a “concrete” category where injective/surjective/kernel/image/exact/etc. agree
with your previous notions. A proof by this means is sometimes called diagramchasing.
Remark 63.4.3 — The “small” condition is a technical obstruction that requires the
objects 𝒜 to actually form a set. I’ll ignore this distinction, because one can almost
always work around it by doing enough set-theoretic technicalities.
For example, let’s prove:
Lemma 63.4.4 (Short five lemma) In an abelian category, consider the commutative diagram
and assume the top and bottom rows are exact. If α and γ are isomorphisms, then so is
β.
Proof.We prove that β is epic (with a similar proof to get monic). By the embedding
theorem we can treat the category as R-modules over some R. This lets us do a
so-called “diagram chase” where we move elements around the picture, using the concrete
interpretation of our category as R-modules.
Let b′ be an element of B′. Then q′(b′) ∈ C′, and since γ is surjective, we have a c such that
γ(c) = b′, and finally a b ∈ B such that q(b) = c. Picture:
Now, it is not necessarily the case that β(b) = b′. However, since the diagram commutes we at
least have that
so b′−β(b) ∈ Kerq′ = Imp′, and there is an a′∈ A′ such that p′(a′) = b′−β(b); use α now
to lift it to a ∈ A. Picture:
Then, we have
so b′∈ Imβ which completes the proof that β′ is surjective. □
63.5Breaking long exact sequences
Prototypical example for this section:First isomorphism theorem.
In fact, it turns out that any exact sequence breaks into short exact sequences. This relies
on:
Proposition 63.5.1 (“First isomorphism theorem” in abelian categories) Let AB be an
arrow of an abelian category. Then there is an exact sequence
Example 63.5.2 Let’s analyze this theorem in our two examples of abelian categories:
(a)
In the category of abelian groups, this is basically the first isomorphism theorem.
(b)
In the category Vectk, this amounts to the rank-nullity theorem, ?? .
Thus, any exact sequence can be broken into short exact sequences, as
where Ck = imfk−1 = kerfk for every k.
63.6A few harder problems to think about
Problem 63A (Four lemma).In an abelian category, consider the commutative diagram
where the first and second rows are exact. Prove that if α is epic, and β and δ are monic, then
γ is epic.
Solution. Let c ∈ C with γ(c) = 0. We show c = 0. This proceeds in a diagram chase:
Note that 0 = r′(γ(c)) = δ(r(c)), and since δ is injective, it follows that r(c) = 0.
Since the top row is exact, it follows c = q(b) for some b ∈ B.
Then q′(β(b)) = 0, so if we let b′ = β(b), then b′∈ ker(q′). As the bottom row is
exact, there exists a′ with p′(a′) = b′.
Since α is injective, there is a ∈ A with α(a′) = a.
Since β is injective, it follows that p(a) = b.
Since the top row is exact, and b is in the image of p, it follows that 0 = q(b) = c
as needed.
□
¡++¿
Problem 63B (Five lemma).In an abelian category, consider the
commutative diagram
where the first and second rows are exact. Prove that if α is epic, 𝜀 is monic, and β, δ are
isomorphisms, then γ is an isomorphism as well. Thus this is a stronger version of the short
five lemma.
Problem 63C⋆ (Snake lemma).In an abelian category, consider the
diagram